Euclid's Elements of Geometry

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κατὰ τὸ Β, καὶ ἀπὸ τῆς ἁφῆς ἦκται τῇ ἐφαπτομένῃ πρὸς and CB have been joined. ELEMENTS BOOK 3
ὀρθὰς ἡ ΒΑ, ἐπὶ τῆς ΒΑ ἄρα τὸ κέντρον ἐστὶ τοῦ ΑΒΓΔ And since some straight-line EF touches the circle
κύκλου. ἡ ΒΑ ἄρα διάμετός ἐστι τοῦ ΑΒΓΔ κύκλου· ἡ ἄρα ABCD at point B, and BA has been drawn from the
ὑπὸ ΑΔΒ γωνία ἐν ἡμικυκλίῳ οὖσα ὀρθή ἐστιν. λοιπαὶ ἄρα point of contact, at right-angles to the tangent, the center
αἱ ὑπὸ ΒΑΔ, ΑΒΔ μιᾷ ὀρθῇ ἴσαι εἰσίν. ἐστὶ δὲ καὶ ἡ ὑπὸ of circle ABCD is thus on BA [Prop. 3.19]. Thus, BA
ΑΒΖ ὀρθή· ἡ ἄρα ὑπὸ ΑΒΖ ἴση ἐστὶ ταῖς ὑπὸ ΒΑΔ, ΑΒΔ. is a diameter of circle ABCD. Thus, angle ADB, being
κοινὴ ἀφῃρήσθω ἡ ὑπὸ ΑΒΔ· λοιπὴ ἄρα ἡ ὑπὸ ΔΒΖ γωνία in a semi-circle, is a right-angle [Prop. 3.31]. Thus, the
ἴση ἐστὶ τῇ ἐν τῷ ἐναλλὰξ τμήματι τοῦ κύκλου γωνίᾳ τῇ remaining angles (of triangle ADB) BAD and ABD are
ὑπὸ ΒΑΔ. καὶ ἐπεὶ ἐν κύκλῳ τετράπλευρόν ἐστι τὸ ΑΒΓΔ, equal to one right-angle [Prop. 1.32]. And ABF is also a
αἱ ἀπεναντίον αὐτοῦ γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν. εἰσὶ δὲ right-angle. Thus, ABF is equal to BAD and ABD. Let
καὶ αἱ ὑπὸ ΔΒΖ, ΔΒΕ δυσὶν ὀρθαῖς ἴσαι· αἱ ἄρα ὑπὸ ΔΒΖ, ABD have been subtracted from both. Thus, the remain-
ΔΒΕ ταῖς ὑπὸ ΒΑΔ, ΒΓΔ ἴσαι εἰσίν, ὧν ἡ ὑπὸ ΒΑΔ τῇ ὑπὸ ing angle DBF is equal to the angle BAD in the alternate
ΔΒΖ ἐδείχθη ἴση· λοιπὴ ἄρα ἡ ὑπὸ ΔΒΕ τῇ ἐν τῷ ἐναλλὰξ segment of the circle. And since ABCD is a quadrilateral
τοῦ κύκλου τμήματι τῷ ΔΓΒ τῇ ὑπὸ ΔΓΒ γωνίᾳ ἐστὶν ἴση. in a circle, (the sum of) its opposite angles is equal to
᾿Εὰν ἄρα κύκλου ἐφάπτηταί τις εὐθεῖα, ἀπὸ δὲ τῆς ἁφῆς two right-angles [Prop. 3.22]. And DBF and DBE is
εἰς τὸν κύκλον διαχθῇ τις εὐθεῖα τέμνουσα τὸν κύκλον, ἃς also equal to two right-angles [Prop. 1.13]. Thus, DBF
ποιεῖ γωνίας πρὸς τῇ ἐφαπτομένῃ, ἴσαι ἔσονται ταῖς ἐν τοῖς and DBE is equal to BAD and BCD, of which BAD
ἐναλλὰξ τοῦ κύκλου τμήμασι γωνίαις· ὅπερ ἔδει δεῖξαι. was shown (to be) equal to DBF. Thus, the remaining
(angle) DBE is equal to the angle DCB in the alternate
lgþ (other) straight-line is drawn across, from the point of
segment DCB of the circle.
Thus, if some straight-line touches a circle, and some
contact into the circle, cutting the circle (in two), then
those angles the (straight-line) makes with the tangent
will be equal to the angles in the alternate segments of
the circle. (Which is) the very thing it was required to
show.

.
Proposition 33
᾿Επὶ τῆς δοθείσης εὐθείας γράψαι τμῆμα κύκλου δεχόμε- To draw a segment of a circle, accepting an angle
νον γωνίαν ἴσην τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ. equal to a given rectilinear angle, on a given straight-line.


C C C
Γ Γ Γ A A A
Α Α Α D D D
∆ ∆ ∆ H
Θ
F
Ζ
Ζ Ε F G E F G
Η Ζ Η
Β B
Β B
Ε Β Ε E B E
῎Εστω ἡ δοθεῖσα εὐθεῖα ἡ ΑΒ, ἡ δὲ δοθεῖσα γωνία Let AB be the given straight-line, and C the given
εὐθύγραμμος ἡ πρὸς τῷ Γ· δεῖ δὴ ἐπὶ τῆς δοθείσης εὐθείας rectilinear angle. So it is required to draw a segment
τῆς ΑΒ γράψαι τμῆμα κύκλου δεχόμενον γωνίαν ἴσην τῇ of a circle, accepting an angle equal to C, on the given
πρὸς τῷ Γ. straight-line AB.
῾Η δὴ πρὸς τῷ Γ [γωνία] ἤτοι ὀξεῖά ἐστιν ἢ ὀρθὴ ἢ So the [angle] C is surely either acute, a right-angle,
ἀμβλεῖα· ἔστω πρότερον ὀξεῖα, καὶ ὡς ἐπὶ τῆς πρώτης κα- or obtuse. First of all, let it be acute. And, as in the first
ταγραφῆς συνεστάτω πρὸς τῇ ΑΒ εὐθείᾳ καὶ τῷ Α σημείῳ diagram (from the left), let (angle) BAD, equal to angle
τῇ πρὸς τῷ Γ γωνίᾳ ἴση ἡ ὑπὸ ΒΑΔ· ὀξεῖα ἄρα ἐστὶ καὶ ἡ C, have been constructed on the straight-line AB, at the
ὑπὸ ΒΑΔ. ἤχθω τῇ ΔΑ πρὸς ὀρθὰς ἡ ΑΕ, καὶ τετμήσθω point A (on it) [Prop. 1.23]. Thus, BAD is also acute. Let
ἡ ΑΒ δίχα κατὰ τὸ Ζ, καὶ ἤχθω ἀπὸ τοῦ Ζ σημείου τῇ ΑΒ AE have been drawn, at right-angles to DA [Prop. 1.11].


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πρὸς ὀρθὰς ἡ ΖΗ, καὶ ἐπεζεύχθω ἡ ΗΒ. And let AB have been cut in half at F [Prop. 1.10]. And
Καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΖ τῇ ΖΒ, κοινὴ δὲ ἡ ΖΗ, δύο δὴ let FG have been drawn from point F, at right-angles to
αἱ ΑΖ, ΖΗ δύο ταῖς ΒΖ, ΖΗ ἴσαι εἰσίν· καὶ γωνία ἡ ὑπὸ AB [Prop. 1.11]. And let GB have been joined.
ΑΖΗ [γωνίᾳ] τῇ ὑπὸ ΒΖΗ ἴση· βάσις ἄρα ἡ ΑΗ βάσει τῇ And since AF is equal to FB, and FG (is) common,
ΒΗ ἴση ἐστίν. ὁ ἄρα κέντρῳ μὲν τῷ Η διαστήματι δὲ τῷ the two (straight-lines) AF, FG are equal to the two
ΗΑ κύκλος γραφόμενος ἥξει καὶ διὰ τοῦ Β. γεγράφθω καὶ (straight-lines) BF, FG (respectively). And angle AFG
ἔστω ὁ ΑΒΕ, καὶ ἐπεζεύχθω ἡ ΕΒ. ἐπεὶ οὖν ἀπ᾿ ἄκρας τῆς (is) equal to [angle] BFG. Thus, the base AG is equal to
ΑΕ διαμέτρου ἀπὸ τοῦ Α τῇ ΑΕ πρὸς ὀρθάς ἐστιν ἡ ΑΔ, the base BG [Prop. 1.4]. Thus, the circle drawn with
ἡ ΑΔ ἄρα ἐφάπτεται τοῦ ΑΒΕ κύκλου· ἐπεὶ οὖν κύκλου center G, and radius GA, will also go through B (as
τοῦ ΑΒΕ ἐφάπτεταί τις εὐθεῖα ἡ ΑΔ, καὶ ἀπὸ τῆς κατὰ τὸ well as A). Let it have been drawn, and let it be (de-
Α ἁφῆς εἰς τὸν ΑΒΕ κύκλον διῆκταί τις εὐθεῖα ἡ ΑΒ, ἡ noted) ABE. And let EB have been joined. Therefore,
ἄρα ὑπὸ ΔΑΒ γωνία ἴση ἐστὶ τῇ ἐν τῷ ἐναλλὰξ τοῦ κύκλου since AD is at the extremity of diameter AE, (namely,
τμήματι γωνίᾳ τῇ ὑπὸ ΑΕΒ. ἀλλ᾿ ἡ ὑπὸ ΔΑΒ τῇ πρὸς τῷ Γ point) A, at right-angles to AE, the (straight-line) AD
ἐστιν ἴση· καὶ ἡ πρὸς τῷ Γ ἄρα γωνία ἴση ἐστὶ τῇ ὑπὸ ΑΕΒ. thus touches the circle ABE [Prop. 3.16 corr.]. There-
᾿Επὶ τῆς δοθείσης ἄρα εὐθείας τῆς ΑΒ τμῆμα κύκλου fore, since some straight-line AD touches the circle ABE,
γέγραπται τὸ ΑΕΒ δεχόμενον γωνίαν τὴν ὑπὸ ΑΕΒ ἴσην and some (other) straight-line AB has been drawn across
τῇ δοθείσῃ τῇ πρὸς τῷ Γ. from the point of contact A into circle ABE, angle DAB
᾿Αλλὰ δὴ ὀρθὴ ἔστω ἡ πρὸς τῷ Γ· καὶ δέον πάλιν ἔστω is thus equal to the angle AEB in the alternate segment
ἐπὶ τῆς ΑΒ γράψαι τμῆμα κύκλου δεχόμενον γωνίαν ἴσην τῇ of the circle [Prop. 3.32]. But, DAB is equal to C. Thus,
πρὸς τῷ Γ ὀρθῇ [γωνίᾳ]. συνεστάτω [πάλιν] τῇ πρὸς τῷ Γ angle C is also equal to AEB.
ὀρθῇ γωνίᾳ ἴση ἡ ὑπὸ ΒΑΔ, ὡς ἔχει ἐπὶ τῆς δευτέρας κατα- Thus, a segment AEB of a circle, accepting the angle
γραφῆς, καὶ τετμήσθω ἡ ΑΒ δίχα κατὰ τὸ Ζ, καὶ κέντρῳ τῷ AEB (which is) equal to the given (angle) C, has been
Ζ, διαστήματι δὲ ὁποτέρῳ τῶν ΖΑ, ΖΒ, κύκλος γεγράφθω drawn on the given straight-line AB.
ὁ ΑΕΒ. And so let C be a right-angle. And let it again be
᾿Εφάπτεται ἄρα ἡ ΑΔ εὐθεῖα τοῦ ΑΒΕ κύκλου διὰ τὸ necessary to draw a segment of a circle on AB, accepting
ὀρθὴν εἶναι τὴν πρὸς τῷ Α γωνίαν. καὶ ἴση ἐστὶν ἡ ὑπὸ an angle equal to the right-[angle] C. Let the (angle)
ΒΑΔ γωνία τῇ ἐν τῷ ΑΕΒ τμήματι· ὀρθὴ γὰρ καὶ αὐτὴ ἐν BAD [again] have been constructed, equal to the right-
ἡμικυκλίῳ οὖσα. ἀλλὰ καὶ ἡ ὑπὸ ΒΑΔ τῇ πρὸς τῷ Γ ἴση angle C [Prop. 1.23], as in the second diagram (from the
ἐστίν. καὶ ἡ ἐν τῷ ΑΕΒ ἄρα ἴση ἐστὶ τῇ πρὸς τῷ Γ. left). And let AB have been cut in half at F [Prop. 1.10].
Γέγραπται ἄρα πάλιν ἐπὶ τῆς ΑΒ τμῆμα κύκλου τὸ ΑΕΒ And let the circle AEB have been drawn with center F,
δεχόμενον γωνίαν ἴσην τῇ πρὸς τῷ Γ. and radius either FA or FB.
᾿Αλλὰ δὴ ἡ πρὸς τῷ Γ ἀμβλεῖα ἔστω· καὶ συνεστάτω Thus, the straight-line AD touches the circle ABE, on
αὐτῇ ἴση πρὸς τῇ ΑΒ εὐθείᾳ καὶ τῷ Α σημείῳ ἡ ὑπὸ ΒΑΔ, account of the angle at A being a right-angle [Prop. 3.16
ὡς ἔχει ἐπὶ τῆς τρίτης καταγραφῆς, καὶ τῇ ΑΔ πρὸς ὀρθὰς corr.]. And angle BAD is equal to the angle in segment
ἤχθω ἡ ΑΕ, καὶ τετμήσθω πάλιν ἡ ΑΒ δίχα κατὰ τὸ Ζ, καὶ AEB. For (the latter angle), being in a semi-circle, is also
τῇ ΑΒ πρὸς ὀρθὰς ἤχθω ἡ ΖΗ, καὶ ἐπεζεύχθω ἡ ΗΒ. a right-angle [Prop. 3.31]. But, BAD is also equal to C.
Καὶ ἐπεὶ πάλιν ἴση ἐστὶν ἡ ΑΖ τῇ ΖΒ, καὶ κοινὴ ἡ ΖΗ, Thus, the (angle) in (segment) AEB is also equal to C.
δύο δὴ αἱ ΑΖ, ΖΗ δύο ταῖς ΒΖ, ΖΗ ἴσαι εἰσίν· καὶ γωνία ἡ Thus, a segment AEB of a circle, accepting an angle
ὑπὸ ΑΖΗ γωνίᾳ τῇ ὑπὸ ΒΖΗ ἴση· βάσις ἄρα ἡ ΑΗ βάσει equal to C, has again been drawn on AB.
τῇ ΒΗ ἴση ἐστίν· ὁ ἄρα κέντρῳ μὲν τῷ Η διαστήματι δὲ τῷ And so let (angle) C be obtuse. And let (angle) BAD,
ΗΑ κύκλος γραφόμενος ἥξει καὶ διὰ τοῦ Β. ἐρχέσθω ὡς ὁ equal to (C), have been constructed on the straight-line
ΑΕΒ. καὶ ἐπεὶ τῇ ΑΕ διαμέτρῳ ἀπ᾿ ἄκρας πρὸς ὀρθάς ἐστιν AB, at the point A (on it) [Prop. 1.23], as in the third
ἡ ΑΔ, ἡ ΑΔ ἄρα ἐφάπτεται τοῦ ΑΕΒ κύκλου. καὶ ἀπὸ τῆς diagram (from the left). And let AE have been drawn, at
κατὰ τὸ Α ἐπαφῆς διῆκται ἡ ΑΒ· ἡ ἄρα ὑπὸ ΒΑΔ γωνία right-angles to AD [Prop. 1.11]. And let AB have again
ἴση ἐστὶ τῇ ἐν τῷ ἐναλλὰξ τοῦ κύκλου τμήματι τῷ ΑΘΒ been cut in half at F [Prop. 1.10]. And let FG have been
συνισταμένῃ γωνίᾳ. ἀλλ᾿ ἡ ὑπὸ ΒΑΔ γωνία τῇ πρὸς τῷ Γ drawn, at right-angles to AB [Prop. 1.10]. And let GB
ἴση ἐστίν. καὶ ἡ ἐν τῷ ΑΘΒ ἄρα τμήματι γωνία ἴση ἐστὶ τῇ have been joined.
πρὸς τῷ Γ. And again, since AF is equal to FB, and FG (is)
᾿Επὶ τῆς ἄρα δοθείσης εὐθείας τῆς ΑΒ γέγραπται τμῆμα common, the two (straight-lines) AF, FG are equal to
κύκλου τὸ ΑΘΒ δεχόμενον γωνίαν ἴσην τῇ πρὸς τῷ Γ· ὅπερ the two (straight-lines) BF, FG (respectively). And an-
ἔδει ποιῆσαι. gle AFG (is) equal to angle BFG. Thus, the base AG is


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equal to the base BG [Prop. 1.4]. Thus, a circle of center
G, and radius GA, being drawn, will also go through B
(as well as A). Let it go like AEB (in the third diagram
from the left). And since AD is at right-angles to the di-
ameter AE, at its extremity, AD thus touches circle AEB
[Prop. 3.16 corr.]. And AB has been drawn across (the
ldþ AHB of the circle [Prop. 3.32]. But, angle BAD is equal
circle) from the point of contact A. Thus, angle BAD is
equal to the angle constructed in the alternate segment

to C. Thus, the angle in segment AHB is also equal to
C.
Thus, a segment AHB of a circle, accepting an angle
equal to C, has been drawn on the given straight-line AB.
(Which is) the very thing it was required to do.

.
Proposition 34
᾿Απὸ τοῦ δοθέντος κύκλου τμῆμα ἀφελεῖν δεχόμενον To cut off a segment, accepting an angle equal to a
γωνίαν ἴσην τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ. given rectilinear angle, from a given circle.
Γ C
Ζ F





Β B


∆ D

Ε Α E A
῎Εστω ὁ δοθεὶς κύκλος ὁ ΑΒΓ, ἡ δὲ δοθεῖσα γωνία Let ABC be the given circle, and D the given rectilin-
εὐθύγραμμος ἡ πρὸς τῷ Δ· δεῖ δὴ ἀπὸ τοῦ ΑΒΓ κύκλου ear angle. So it is required to cut off a segment, accepting
τμῆμα ἀφελεῖν δεχόμενον γωνίαν ἴσην τῇ δοθείσῃ γωνίᾳ an angle equal to the given rectilinear angle D, from the
εὐθυγράμμῳ τῇ πρὸς τῷ Δ. given circle ABC.
῎Ηχθω τοῦ ΑΒΓ ἐφαπτομένη ἡ ΕΖ κατὰ τὸ Β σημεῖον, Let EF have been drawn touching ABC at point B. †
καὶ συνεστάτω πρὸς τῇ ΖΒ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ And let (angle) FBC, equal to angle D, have been con-
τῷ Β τῇ πρὸς τῷ Δ γωνίᾳ ἴση ἡ ὑπὸ ΖΒΓ. structed on the straight-line FB, at the point B on it
᾿Επεὶ οὖν κύκλου τοῦ ΑΒΓ ἐφάπτεταί τις εὐθεῖα ἡ ΕΖ, [Prop. 1.23].
καὶ ἀπὸ τῆς κατὰ τὸ Β ἐπαφῆς διῆκται ἡ ΒΓ, ἡ ὑπὸ ΖΒΓ ἄρα Therefore, since some straight-line EF touches the
γωνία ἴση ἐστὶ τῇ ἐν τῷ ΒΑΓ ἐναλλὰξ τμήματι συνισταμένῃ circle ABC, and BC has been drawn across (the circle)
γωνίᾳ. ἀλλ᾿ ἡ ὑπὸ ΖΒΓ τῇ πρὸς τῷ Δ ἐστιν ἴση· καὶ ἡ ἐν from the point of contact B, angle FBC is thus equal
τῷ ΒΑΓ ἄρα τμήματι ἴση ἐστὶ τῇ πρὸς τῷ Δ [γωνίᾳ]. to the angle constructed in the alternate segment BAC
᾿Απὸ τοῦ δοθέντος ἄρα κύκλου τοῦ ΑΒΓ τμῆμα ἀφῄρηται [Prop. 1.32]. But, FBC is equal to D. Thus, the (angle)
τὸ ΒΑΓ δεχόμενον γωνίαν ἴσην τῇ δοθείσῃ γωνίᾳ εὐθυγράμ- in the segment BAC is also equal to [angle] D.
μῳ τῇ πρὸς τῷ Δ· ὅπερ ἔδει ποιῆσαι. Thus, the segment BAC, accepting an angle equal to
the given rectilinear angle D, has been cut off from the
given circle ABC. (Which is) the very thing it was re-
quired to do.

† Presumably, by finding the center of ABC [Prop. 3.1], drawing a straight-line between the center and point B, and then drawing EF through



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point B, at right-angles to the aforementioned straight-line [Prop. 1.11]. ELEMENTS BOOK 3
Proposition 35
.
᾿Εὰν ἐν κύκλῳ δύο εὐθεῖαι τέμνωσιν ἀλλήλας, τὸ ὑπὸ If two straight-lines in a circle cut one another then
τῶν τῆς μιᾶς τμημάτων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ the rectangle contained by the pieces of one is equal to
τῷ ὑπὸ τῶν τῆς ἑτέρας τμημάτων περιεχομένῳ ὀρθογωνίῳ. the rectangle contained by the pieces of the other.
Α A
∆ D
Α A
Ε Ζ E F
Β ∆ B D
Η Θ G H
Ε E
Γ Γ C C
Β B
᾿Εν γὰρ κύκλῳ τῷ ΑΒΓΔ δύο εὐθεῖαι αἱ ΑΓ, ΒΔ For let the two straight-lines AC and BD, in the circle
τεμνέτωσαν ἀλλήλας κατὰ τὸ Ε σημεῖον· λέγω, ὅτι τὸ ὑπὸ ABCD, cut one another at point E. I say that the rect-
τῶν ΑΕ, ΕΓ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ angle contained by AE and EC is equal to the rectangle
τῶν ΔΕ, ΕΒ περιεχομένῳ ὀρθογωνίῳ. contained by DE and EB.
Εἰ μὲν οὖν αἱ ΑΓ, ΒΔ διὰ τοῦ κέντρου εἰσὶν ὥστε τὸ Ε In fact, if AC and BD are through the center (as in
κέντρον εἶναι τοῦ ΑΒΓΔ κύκλου, φανερόν, ὅτι ἴσων οὐσῶν the first diagram from the left), so that E is the center of
τῶν ΑΕ, ΕΓ, ΔΕ, ΕΒ καὶ τὸ ὑπὸ τῶν ΑΕ, ΕΓ περιεχόμενον circle ABCD, then (it is) clear that, AE, EC, DE, and
ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν ΔΕ, ΕΒ περιεχομένῳ EB being equal, the rectangle contained by AE and EC
ὀρθογωνίῳ. is also equal to the rectangle contained by DE and EB.
Μὴ ἔστωσαν δὴ αἱ ΑΓ, ΔΒ διὰ τοῦ κέντρου, καὶ So let AC and DB not be though the center (as in
εἰλήφθω τὸ κέντρον τοῦ ΑΒΓΔ, καὶ ἔστω τὸ Ζ, καὶ ἀπὸ the second diagram from the left), and let the center of
τοῦ Ζ ἐπὶ τὰς ΑΓ, ΔΒ εὐθείας κάθετοι ἤχθωσαν αἱ ΖΗ, ABCD have been found [Prop. 3.1], and let it be (at) F.
ΖΘ, καὶ ἐπεζεύχθωσαν αἱ ΖΒ, ΖΓ, ΖΕ. And let FG and FH have been drawn from F, perpen-
Καὶ ἐπεὶ εὐθεῖά τις διὰ τοῦ κέντρου ἡ ΗΖ εὐθεῖάν τινα dicular to the straight-lines AC and DB (respectively)
μὴ διὰ τοῦ κέντρου τὴν ΑΓ πρὸς ὀρθὰς τέμνει, καὶ δίχα [Prop. 1.12]. And let FB, FC, and FE have been joined.
αὐτὴν τέμνει· ἴση ἄρα ἡ ΑΗ τῇ ΗΓ. ἐπεὶ οὖν εὐθεῖα ἡ ΑΓ And since some straight-line, GF, through the center,
τέτμηται εἰς μὲν ἴσα κατὰ τὸ Η, εἰς δὲ ἄνισα κατὰ τὸ Ε, τὸ cuts at right-angles some (other) straight-line, AC, not
ἄρα ὑπὸ τῶν ΑΕ, ΕΓ περιεχόμενον ὀρθογώνιον μετὰ τοῦ through the center, then it also cuts it in half [Prop. 3.3].
ἀπὸ τῆς ΕΗ τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς ΗΓ· [κοινὸν] Thus, AG (is) equal to GC. Therefore, since the straight-
προσκείσθω τὸ ἀπὸ τῆς ΗΖ· τὸ ἄρα ὑπὸ τῶν ΑΕ, ΕΓ μετὰ line AC is cut equally at G, and unequally at E, the
τῶν ἀπὸ τῶν ΗΕ, ΗΖ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΓΗ, ΗΖ. ἀλλὰ rectangle contained by AE and EC plus the square on
τοῖς μὲν ἀπὸ τῶν ΕΗ, ΗΖ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΖΕ, τοὶς EG is thus equal to the (square) on GC [Prop. 2.5]. Let
δὲ ἀπὸ τῶν ΓΗ, ΗΖ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΖΓ· τὸ ἄρα ὑπὸ the (square) on GF have been added [to both]. Thus,
τῶν ΑΕ, ΕΓ μετὰ τοῦ ἀπὸ τῆς ΖΕ ἴσον ἐστὶ τῷ ἀπὸ τῆς the (rectangle contained) by AE and EC plus the (sum
ΖΓ. ἴση δὲ ἡ ΖΓ τῇ ΖΒ· τὸ ἄρα ὑπὸ τῶν ΑΕ, ΕΓ μετὰ τοῦ of the squares) on GE and GF is equal to the (sum of
ἀπὸ τῆς ΕΖ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΖΒ. διὰ τὰ αὐτὰ δὴ καὶ the squares) on CG and GF. But, the (square) on FE
τὸ ὑπὸ τῶν ΔΕ, ΕΒ μετὰ τοῦ ἀπὸ τῆς ΖΕ ἰσον ἐστὶ τῷ ἀπὸ is equal to the (sum of the squares) on EG and GF
τῆς ΖΒ. ἐδείχθη δὲ καὶ τὸ ὑπὸ τῶν ΑΕ, ΕΓ μετὰ τοῦ ἀπὸ [Prop. 1.47], and the (square) on FC is equal to the (sum
τῆς ΖΕ ἴσον τῷ ἀπὸ τῆς ΖΒ· τὸ ἄρα ὑπὸ τῶν ΑΕ, ΕΓ μετὰ of the squares) on CG and GF [Prop. 1.47]. Thus, the
τοῦ ἀπὸ τῆς ΖΕ ἴσον ἐστὶ τῷ ὑπὸ τῶν ΔΕ, ΕΒ μετὰ τοῦ (rectangle contained) by AE and EC plus the (square)
ἀπὸ τῆς ΖΕ. κοινὸν ἀφῇρήσθω τὸ ἀπὸ τῆς ΖΕ· λοιπὸν ἄρα on FE is equal to the (square) on FC. And FC (is)
τὸ ὑπὸ τῶν ΑΕ, ΕΓ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ equal to FB. Thus, the (rectangle contained) by AE
ὑπὸ τῶν ΔΕ, ΕΒ περιεχομένῳ ὀρθογωνίῳ. and EC plus the (square) on FE is equal to the (square)
᾿Εὰν ἄρα ἐν κύκλῳ εὐθεῖαι δύο τέμνωσιν ἀλλήλας, τὸ on FB. So, for the same (reasons), the (rectangle con-
ὑπὸ τῶν τῆς μιᾶς τμημάτων περιεχόμενον ὀρθογώνιον ἴσον tained) by DE and EB plus the (square) on FE is equal


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ἐστὶ τῷ ὑπὸ τῶν τῆς ἑτέρας τμημάτων περιεχομένῳ ὀρθο- to the (square) on FB. And the (rectangle contained)
γωνίῳ· ὅπερ ἔδει δεῖξαι. by AE and EC plus the (square) on FE was also shown
(to be) equal to the (square) on FB. Thus, the (rect-
angle contained) by AE and EC plus the (square) on
FE is equal to the (rectangle contained) by DE and EB
lþ tained by AE and EC is equal to the rectangle contained
plus the (square) on FE. Let the (square) on FE have
been taken from both. Thus, the remaining rectangle con-

by DE and EB.
Thus, if two straight-lines in a circle cut one another
then the rectangle contained by the pieces of one is equal
to the rectangle contained by the pieces of the other.
(Which is) the very thing it was required to show.
Proposition 36
.
᾿Εὰν κύκλου ληφθῇ τι σημεῖον ἐκτός, καὶ ἀπ᾿ αὐτοῦ If some point is taken outside a circle, and two
πρὸς τὸν κύκλον προσπίπτωσι δύο εὐθεῖαι, καὶ ἡ μὲν αὐτῶν straight-lines radiate from it towards the circle, and (one)
τέμνῃ τὸν κύκλον, ἡ δὲ ἐφάπτηται, ἔσται τὸ ὑπὸ ὅλης τῆς of them cuts the circle, and the (other) touches (it), then
τεμνούσης καὶ τῆς ἐκτὸς ἀπολαμβανομένης μεταξὺ τοῦ τε the (rectangle contained) by the whole (straight-line)
σημείου καὶ τῆς κυρτῆς περιφερείας ἴσον τῷ ἀπὸ τῆς ἐφα- cutting (the circle), and the (part of it) cut off outside
πτομένης τετραγώνῳ. (the circle), between the point and the convex circumfer-
ence, will be equal to the square on the tangent (line).
Α A


Ε E
Β Ζ B F
Α A
Γ C
Ζ F
Γ ∆ Β C D B



∆ D
Κύκλου γὰρ τοῦ ΑΒΓ εἰλήφθω τι σημεῖον ἐκτὸς τὸ Δ, For let some point D have been taken outside circle
καὶ ἀπὸ τοῦ Δ πρὸς τὸν ΑΒΓ κύκλον προσπιπτέτωσαν δύο ABC, and let two straight-lines, DC[A] and DB, radi-
εὐθεῖαι αἱ ΔΓ[Α], ΔΒ· καὶ ἡ μὲν ΔΓΑ τεμνέτω τὸν ΑΒΓ ate from D towards circle ABC. And let DCA cut circle
κύκλον, ἡ δὲ ΒΔ ἐφαπτέσθω· λέγω, ὅτι τὸ ὑπὸ τῶν ΑΔ, ABC, and let BD touch (it). I say that the rectangle
ΔΓ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ἀπὸ τῆς ΔΒ contained by AD and DC is equal to the square on DB.
τετραγώνῳ. [D]CA is surely either through the center, or not. Let
῾Η ἄρα [Δ]ΓΑ ἤτοι διὰ τοῦ κέντρου ἐστὶν ἢ οὔ. ἔστω it first of all be through the center, and let F be the cen-
πρότερον διὰ τοῦ κέντρου, καὶ ἔστω τὸ Ζ κέντρον τοῦ ΑΒΓ ter of circle ABC, and let FB have been joined. Thus,
κύκλου, καὶ ἐπεζεύχθω ἡ ΖΒ· ὀρθὴ ἄρα ἐστὶν ἡ ὑπὸ ΖΒΔ. (angle) FBD is a right-angle [Prop. 3.18]. And since
καὶ ἐπεὶ εὐθεῖα ἡ ΑΓ δίχα τέτμηται κατὰ τὸ Ζ, πρόσκειται straight-line AC is cut in half at F, let CD have been
δὲ αὐτῇ ἡ ΓΔ, τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τοῦ ἀπὸ τῆς added to it. Thus, the (rectangle contained) by AD and
ΖΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΖΔ. ἴση δὲ ἡ ΖΓ τῇ ΖΒ· τὸ ἄρα DC plus the (square) on FC is equal to the (square) on
ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τοῦ ἀπὸ τῆς ΖΒ ἴσον ἐστὶ τῷ ἀπὸ FD [Prop. 2.6]. And FC (is) equal to FB. Thus, the
τὴς ΖΔ. τῷ δὲ ἀπὸ τῆς ΖΔ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΖΒ, ΒΔ· (rectangle contained) by AD and DC plus the (square)
τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τοῦ ἀπὸ τῆς ΖΒ ἴσον ἐστὶ on FB is equal to the (square) on FD. And the (square)
τοῖς ἀπὸ τῶν ΖΒ, ΒΔ. κοινὸν ἀφῃρήσθω τὸ ἀπὸ τῆς ΖΒ· on FD is equal to the (sum of the squares) on FB and
λοιπὸν ἄρα τὸ ὑπὸ τῶν ΑΔ, ΔΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΔΒ BD [Prop. 1.47]. Thus, the (rectangle contained) by AD


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ἐφαπτομένης. and DC plus the (square) on FB is equal to the (sum
᾿Αλλὰ δὴ ἡ ΔΓΑ μὴ ἔστω διὰ τοῦ κέντρου τοῦ ΑΒΓ of the squares) on FB and BD. Let the (square) on
κύκλου, καὶ εἰλήφθω τὸ κέντρον τὸ Ε, καὶ ἀπὸ τοῦ Ε ἐπὶ FB have been subtracted from both. Thus, the remain-
τὴν ΑΓ κάθετος ἤχθω ἡ ΕΖ, καὶ ἐπεζεύχθωσαν αἱ ΕΒ, ΕΓ, ing (rectangle contained) by AD and DC is equal to the
ΕΔ· ὀρθὴ ἄρα ἐστὶν ἡ ὑπὸ ΕΒΔ. καὶ ἐπεὶ εὐθεῖά τις διὰ τοῦ (square) on the tangent DB.
κέντρου ἡ ΕΖ εὐθεῖάν τινα μὴ διὰ τοῦ κέντρου τὴν ΑΓ πρὸς And so let DCA not be through the center of cir-
ὀρθὰς τέμνει, καὶ δίχα αὐτὴν τέμνει· ἡ ΑΖ ἄρα τῇ ΖΓ ἐστιν cle ABC, and let the center E have been found, and
ἴση. καὶ ἐπεὶ εὐθεῖα ἡ ΑΓ τέτμηται δίχα κατὰ τὸ Ζ σημεῖον, let EF have been drawn from E, perpendicular to AC
πρόσκειται δὲ αὐτῇ ἡ ΓΔ, τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τοῦ [Prop. 1.12]. And let EB, EC, and ED have been joined.
ἀπὸ τῆς ΖΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΖΔ. κοινὸν προσκείσθω (Angle) EBD (is) thus a right-angle [Prop. 3.18]. And
τὸ ἀπὸ τῆς ΖΕ· τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τῶν ἀπὸ τῶν since some straight-line, EF, through the center, cuts
ΓΖ, ΖΕ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΖΔ, ΖΕ. τοῖς δὲ ἀπὸ τῶν some (other) straight-line, AC, not through the center,
ΓΖ, ΖΕ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΕΓ· ὀρθὴ γὰρ [ἐστιν] ἡ ὑπὸ at right-angles, it also cuts it in half [Prop. 3.3]. Thus,
ΕΖΓ [γωνία]· τοῖς δὲ ἀπὸ τῶν ΔΖ, ΖΕ ἴσον ἐστὶ τὸ ἀπὸ τῆς AF is equal to FC. And since the straight-line AC is cut
ΕΔ· τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τοῦ ἀπὸ τῆς ΕΓ ἴσον in half at point F, let CD have been added to it. Thus, the
ἐστὶ τῷ ἀπὸ τῆς ΕΔ. ἴση δὲ ἡ ΕΓ τῂ ΕΒ· τὸ ἄρα ὑπὸ τῶν (rectangle contained) by AD and DC plus the (square)
ΑΔ, ΔΓ μετὰ τοῦ ἀπὸ τῆς ΕΒ ἴσον ἐστὶ τῷ ἄπὸ τῆς ΕΔ. on FC is equal to the (square) on FD [Prop. 2.6]. Let
τῷ δὲ ἀπὸ τῆς ΕΔ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΕΒ, ΒΔ· ὀρθὴ γὰρ the (square) on FE have been added to both. Thus, the
ἡ ὑπὸ ΕΒΔ γωνία· τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τοῦ ἀπὸ (rectangle contained) by AD and DC plus the (sum of
τῆς ΕΒ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΕΒ, ΒΔ. κοινὸν ἀφῃρήσθω the squares) on CF and FE is equal to the (sum of the
τὸ ἀπὸ τῆς ΕΒ· λοιπὸν ἄρα τὸ ὑπὸ τῶν ΑΔ, ΔΓ ἴσον ἐστὶ squares) on FD and FE. But the (square) on EC is equal
τῷ ἀπὸ τῆς ΔΒ. to the (sum of the squares) on CF and FE. For [angle]
᾿Εὰν ἄρα κύκλου ληφθῇ τι σημεῖον ἐκτός, καὶ ἀπ᾿ αὐτοῦ EFC [is] a right-angle [Prop. 1.47]. And the (square)
πρὸς τὸν κύκλον προσπίπτωσι δύο εὐθεῖαι, καὶ ἡ μὲν αὐτῶν on ED is equal to the (sum of the squares) on DF and
τέμνῃ τὸν κύκλον, ἡ δὲ ἐφάπτηται, ἔσται τὸ ὑπὸ ὅλης τῆς FE [Prop. 1.47]. Thus, the (rectangle contained) by AD
τεμνούσης καὶ τῆς ἐκτὸς ἀπολαμβανομένης μεταξὺ τοῦ τε and DC plus the (square) on EC is equal to the (square)
σημείου καὶ τῆς κυρτῆς περιφερείας ἴσον τῷ ἀπὸ τῆς ἐφα- on ED. And EC (is) equal to EB. Thus, the (rectan-
πτομένης τετραγώνῳ· ὅπερ ἔδει δεῖξαι. gle contained) by AD and DC plus the (square) on EB
is equal to the (square) on ED. And the (sum of the
squares) on EB and BD is equal to the (square) on ED.
For EBD (is) a right-angle [Prop. 1.47]. Thus, the (rect-
angle contained) by AD and DC plus the (square) on
EB is equal to the (sum of the squares) on EB and BD.
Let the (square) on EB have been subtracted from both.
Thus, the remaining (rectangle contained) by AD and
DC is equal to the (square) on BD.
lzþ of them cuts the circle, and (the other) touches (it), then
Thus, if some point is taken outside a circle, and two
straight-lines radiate from it towards the circle, and (one)

the (rectangle contained) by the whole (straight-line)
cutting (the circle), and the (part of it) cut off outside
(the circle), between the point and the convex circumfer-
ence, will be equal to the square on the tangent (line).
(Which is) the very thing it was required to show.

.
Proposition 37
᾿Εὰν κύκλου ληφθῇ τι σημεῖον ἐκτός, ἀπὸ δὲ τοῦ If some point is taken outside a circle, and two
σημείου πρὸς τὸν κύκλον προσπίπτωσι δύο εὐθεῖαι, καὶ straight-lines radiate from the point towards the circle,
ἡ μὲν αὐτῶν τέμνῃ τὸν κύκλον, ἡ δὲ προσπίπτῃ, ᾖ δὲ τὸ and one of them cuts the circle, and the (other) meets
ὑπὸ [τῆς] ὅλης τῆς τεμνούσης καὶ τῆς ἐκτὸς ἀπολαμβα- (it), and the (rectangle contained) by the whole (straight-


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νομένης μεταξὺ τοῦ τε σημείου καὶ τῆς κυρτῆς περιφερείας line) cutting (the circle), and the (part of it) cut off out-
ἴσον τῷ ἀπὸ τῆς προσπιπτούσης, ἡ προσπίπτουσα ἐφάψεται side (the circle), between the point and the convex cir-
τοῦ κύκλου. cumference, is equal to the (square) on the (straight-line)
meeting (the circle), then the (straight-line) meeting (the
circle) will touch the circle.
∆ D
Ε E




Γ C
Ζ F


Β Α B A

Κύκλου γὰρ τοῦ ΑΒΓ εἰλήφθω τι σημεῖον ἐκτὸς τὸ Δ, For let some point D have been taken outside circle
καὶ ἀπὸ τοῦ Δ πρὸς τὸν ΑΒΓ κύκλον προσπιπτέτωσαν δύο ABC, and let two straight-lines, DCA and DB, radiate
εὐθεῖαι αἱ ΔΓΑ, ΔΒ, καὶ ἡ μὲν ΔΓΑ τεμνέτω τὸν κύκλον, ἡ from D towards circle ABC, and let DCA cut the circle,
δὲ ΔΒ προσπιπτέτω, ἔστω δὲ τὸ ὑπὸ τῶν ΑΔ, ΔΓ ἴσον τῷ and let DB meet (the circle). And let the (rectangle con-
ἀπὸ τῆς ΔΒ. λέγω, ὅτι ἡ ΔΒ ἐφάπτεται τοῦ ΑΒΓ κύκλου. tained) by AD and DC be equal to the (square) on DB.
῎Ηχθω γὰρ τοῦ ΑΒΓ ἐφαπτομένη ἡ ΔΕ, καὶ εἰλήφθω τὸ I say that DB touches circle ABC.
κέντρον τοῦ ΑΒΓ κύκλου, καὶ ἔστω τὸ Ζ, καὶ ἐπεζεύχθωσαν For let DE have been drawn touching ABC [Prop.
αἱ ΖΕ, ΖΒ, ΖΔ. ἡ ἄρα ὑπὸ ΖΕΔ ὀρθή ἐστιν. καὶ ἐπεὶ ἡ ΔΕ 3.17], and let the center of the circle ABC have been
ἐφάπτεται τοῦ ΑΒΓ κύκλου, τέμνει δὲ ἡ ΔΓΑ, τὸ ἄρα ὑπὸ found, and let it be (at) F. And let FE, FB, and FD
τῶν ΑΔ, ΔΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΔΕ. ἦν δὲ καὶ τὸ ὑπὸ have been joined. (Angle) FED is thus a right-angle
τῶν ΑΔ, ΔΓ ἴσον τῷ ἀπὸ τῆς ΔΒ· τὸ ἄρα ἀπὸ τῆς ΔΕ [Prop. 3.18]. And since DE touches circle ABC, and
ἴσον ἐστὶ τῷ ἀπὸ τῆς ΔΒ· ἴση ἄρα ἡ ΔΕ τῇ ΔΒ. ἐστὶ δὲ DCA cuts (it), the (rectangle contained) by AD and DC
καὶ ἡ ΖΕ τῇ ΖΒ ἴση· δύο δὴ αἱ ΔΕ, ΕΖ δύο ταῖς ΔΒ, ΒΖ is thus equal to the (square) on DE [Prop. 3.36]. And the
ἴσαι εἰσίν· καὶ βάσις αὐτῶν κοινὴ ἡ ΖΔ· γωνία ἄρα ἡ ὑπὸ (rectangle contained) by AD and DC was also equal to
ΔΕΖ γωνίᾳ τῇ ὑπὸ ΔΒΖ ἐστιν ἴση. ὀρθὴ δὲ ἡ ὑπὸ ΔΕΖ· the (square) on DB. Thus, the (square) on DE is equal
ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΔΒΖ. καί ἐστιν ἡ ΖΒ ἐκβαλλομένη to the (square) on DB. Thus, DE (is) equal to DB. And
διάμετρος· ἡ δὲ τῇ διαμέτρῳ τοῦ κύκλου πρὸς ὀρθὰς ἀπ᾿ FE is also equal to FB. So the two (straight-lines) DE,
ἄκρας ἀγομένη ἐφάπτεται τοῦ κύκλου· ἡ ΔΒ ἄρα ἐφάπτεται EF are equal to the two (straight-lines) DB, BF (re-
τοῦ ΑΒΓ κύκλου. ὁμοίως δὴ δειχθήσεται, κἂν τὸ κέντρον spectively). And their base, FD, is common. Thus, angle
ἐπὶ τῆς ΑΓ τυγχάνῃ. DEF is equal to angle DBF [Prop. 1.8]. And DEF (is)
᾿Εὰν ἄρα κύκλου ληφθῇ τι σημεῖον ἐκτός, ἀπὸ δὲ τοῦ a right-angle. Thus, DBF (is) also a right-angle. And
σημείου πρὸς τὸν κύκλον προσπίπτωσι δύο εὐθεῖαι, καὶ ἡ FB produced is a diameter, And a (straight-line) drawn
μὲν αὐτῶν τέμνῃ τὸν κύκλον, ἡ δὲ προσπίπτῃ, ᾖ δὲ τὸ ὑπὸ at right-angles to a diameter of a circle, at its extremity,
ὅλης τῆς τεμνούσης καὶ τῆς ἐκτὸς ἀπολαμβανομένης μεταξὺ touches the circle [Prop. 3.16 corr.]. Thus, DB touches
τοῦ τε σημείου καὶ τῆς κυρτῆς περιφερείας ἴσον τῷ ἀπὸ circle ABC. Similarly, (the same thing) can be shown,
τῆς προσπιπτούσης, ἡ προσπίπτουσα ἐφάψεται τοῦ κύκλου· even if the center happens to be on AC.
ὅπερ ἔδει δεῖξαι. Thus, if some point is taken outside a circle, and two
straight-lines radiate from the point towards the circle,
and one of them cuts the circle, and the (other) meets
(it), and the (rectangle contained) by the whole (straight-
line) cutting (the circle), and the (part of it) cut off out-
side (the circle), between the point and the convex cir-
cumference, is equal to the (square) on the (straight-line)
meeting (the circle), then the (straight-line) meeting (the
circle) will touch the circle. (Which is) the very thing it


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Construction of Rectilinear Figures In and

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Definitions
.
αʹ. Σχῆμα εὐθύγραμμον εἰς σχῆμα εὐθύγραμμον ἐγγράφ- 1. A rectilinear figure is said to be inscribed in
εσθαι λέγεται, ὅταν ἑκάστη τῶν τοῦ ἐγγραφομένου σχήματ- a(nother) rectilinear figure when the respective angles
ος γωνιῶν ἑκάστης πλευρᾶς τοῦ, εἰς ὃ ἐγγράφεται, ἅπτηται. of the inscribed figure touch the respective sides of the
βʹ. Σχῆμα δὲ ὁμοίως περὶ σχῆμα περιγράφεσθαι λέγεται, (figure) in which it is inscribed.
ὅταν ἑκάστη πλευρὰ τοῦ περιγραφομένου ἑκάστης γωνίας 2. And, similarly, a (rectilinear) figure is said to be cir-
τοῦ, περὶ ὃ περιγράφεται, ἅπτηται. cumscribed about a(nother rectilinear) figure when the
γʹ. Σχῆμα εὐθύγραμμον εἰς κύκλον ἐγγράφεσθαι λέγεται, respective sides of the circumscribed (figure) touch the
ὅταν ἑκάστη γωνία τοῦ ἐγγραφομένου ἅπτηται τῆς τοῦ respective angles of the (figure) about which it is circum-
κύκλου περιφερείας. scribed.
δʹ. Σχῆμα δὲ εὐθύγραμμον περὶ κύκλον περιγράφε- 3. A rectilinear figure is said to be inscribed in a cir-
σθαι λέγεται, ὅταν ἑκάστη πλευρὰ τοῦ περιγραφομένου cle when each angle of the inscribed (figure) touches the
ἐφάπτηται τῆς τοῦ κύκλου περιφερείας. circumference of the circle.
εʹ. Κύκλος δὲ εἰς σχῆμα ὁμοίως ἐγγράφεσθαι λέγεται, 4. And a rectilinear figure is said to be circumscribed
ὅταν ἡ τοῦ κύκλου περιφέρεια ἑκάστης πλευρᾶς τοῦ, εἰς ὃ about a circle when each side of the circumscribed (fig-
ἐγγράφεται, ἅπτηται. ure) touches the circumference of the circle.
ϛʹ. Κύκλος δὲ περὶ σχῆμα περιγράφεσθαι λέγεται, ὅταν 5. And, similarly, a circle is said to be inscribed in a
aþ touches each angle of the (figure) about which it is cir-
ἡ τοῦ κύκλου περιφέρεια ἑκάστης γωνίας τοῦ, περὶ ὃ πε- (rectilinear) figure when the circumference of the circle
ριγράφεται, ἅπτηται. touches each side of the (figure) in which it is inscribed.
ζʹ. Εὐθεῖα εἰς κύκλον ἐναρμόζεσθαι λέγεται, ὅταν τὰ 6. And a circle is said to be circumscribed about a
πέρατα αὐτῆς ἐπὶ τῆς περιφερείας ᾖ τοῦ κύκλου. rectilinear (figure) when the circumference of the circle

cumscribed.
7. A straight-line is said to be inserted into a circle
when its extemities are on the circumference of the circle.

.
Proposition 1
Εἰς τὸν δοθέντα κύκλον τῇ δοθείσῃ εὐθείᾳ μὴ μείζονι To insert a straight-line equal to a given straight-line
οὔσῃ τῆς τοῦ κύκλου διαμέτρου ἴσην εὐθεῖαν ἐναρμόσαι. into a circle, (the latter straight-line) not being greater
than the diameter of the circle.
∆ D



Α A




Ε E
Β Γ B C




Ζ F
῎Εστω ὁ δοθεὶς κύκλος ὁ ΑΒΓ, ἡ δὲ δοθεῖσα εὐθεῖα μὴ Let ABC be the given circle, and D the given straight-
μείζων τῆς τοῦ κύκλου διαμέτρου ἡ Δ. δεῖ δὴ εἰς τὸν ΑΒΓ line (which is) not greater than the diameter of the cir-
κύκλον τῇ Δ εὐθείᾳ ἴσην εὐθεῖαν ἐναρμόσαι. cle. So it is required to insert a straight-line, equal to the
῎Ηχθω τοῦ ΑΒΓ κύκλου διάμετρος ἡ ΒΓ. εἰ μὲν οὖν ἴση straight-line D, into the circle ABC.
ἐστὶν ἡ ΒΓ τῇ Δ, γεγονὸς ἂν εἴη τὸ ἐπιταχθέν· ἐνήρμοσται Let a diameter BC of circle ABC have been drawn. †


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γὰρ εἰς τὸν ΑΒΓ κύκλον τῇ Δ εὐθείᾳ ἴση ἡ ΒΓ. εἰ δὲ μείζων Therefore, if BC is equal to D then that (which) was
ἐστὶν ἡ ΒΓ τῆς Δ, κείσθω τῇ Δ ἴση ἡ ΓΕ, καὶ κέντρῳ prescribed has taken place. For the (straight-line) BC,
τῷ Γ διαστήματι δὲ τῷ ΓΕ κύκλος γεγράφθω ὁ ΕΑΖ, καὶ equal to the straight-line D, has been inserted into the
ἐπεζεύχθω ἡ ΓΑ. circle ABC. And if BC is greater than D then let CE be
᾿Επεὶ οὖν το Γ σημεῖον κέντρον ἐστὶ τοῦ ΕΑΖ κύκλου, made equal to D [Prop. 1.3], and let the circle EAF have
ἴση ἐστὶν ἡ ΓΑ τῇ ΓΕ. ἀλλὰ τῇ Δ ἡ ΓΕ ἐστιν ἴση· καὶ ἡ Δ been drawn with center C and radius CE. And let CA
ἄρα τῇ ΓΑ ἐστιν ἴση. have been joined.
bþ inserted into the given circle ABC. (Which is) the very
Εἰς ἄρα τὸν δοθέντα κύκλον τὸν ΑΒΓ τῇ δοθείσῃ εὐθείᾳ Therefore, since the point C is the center of circle
τῇ Δ ἴση ἐνήρμοσται ἡ ΓΑ· ὅπερ ἔδει ποιῆσαι. EAF, CA is equal to CE. But, CE is equal to D. Thus,
D is also equal to CA.
Thus, CA, equal to the given straight-line D, has been


thing it was required to do.
† Presumably, by finding the center of the circle [Prop. 3.1], and then drawing a line through it.

Proposition 2
.
Εἰς τὸν δοθέντα κύκλον τῷ δοθέντι τριγώνῳ ἰσογώνιον To inscribe a triangle, equiangular with a given trian-
τρίγωνον ἐγγράψαι. gle, in a given circle.
Β Ε B E





Ζ F
Γ C
Η G
∆ D

Α A

Θ H
῎Εστω ὁ δοθεὶς κύκλος ὁ ΑΒΓ, τὸ δὲ δοθὲν τριγωνον Let ABC be the given circle, and DEF the given tri-
τὸ ΔΕΖ· δεῖ δὴ εἰς τὸν ΑΒΓ κύκλον τῷ ΔΕΖ τριγώνῳ angle. So it is required to inscribe a triangle, equiangular
ἰσογώνιον τρίγωνον ἐγγράψαι. with triangle DEF, in circle ABC.
῎Ηχθω τοῦ ΑΒΓ κύκλου ἐφαπτομένη ἡ ΗΘ κατὰ τὸ Α, Let GH have been drawn touching circle ABC at A. †
καὶ συνεστάτω πρὸς τῇ ΑΘ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ And let (angle) HAC, equal to angle DEF, have been
τῷ Α τῇ ὑπὸ ΔΕΖ γωνίᾳ ἴση ἡ ὑπὸ ΘΑΓ, πρὸς δὲ τῇ ΑΗ constructed on the straight-line AH at the point A on it,
εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α τῇ ὑπὸ ΔΖΕ [γωνίᾳ] and (angle) GAB, equal to [angle] DFE, on the straight-
ἴση ἡ ὑπὸ ΗΑΒ, καὶ ἐπεζεύχθω ἡ ΒΓ. line AG at the point A on it [Prop. 1.23]. And let BC
᾿Επεὶ οὖν κύκλου τοῦ ΑΒΓ ἐφάπτεταί τις εὐθεῖα ἡ ΑΘ, have been joined.
καὶ ἀπὸ τῆς κατὰ τὸ Α ἐπαφῆς εἰς τὸν κύκλον διῆκται εὐθεῖα Therefore, since some straight-line AH touches the
ἡ ΑΓ, ἡ ἄρα ὑπὸ ΘΑΓ ἴση ἐστὶ τῇ ἐν τῷ ἐναλλὰξ τοῦ κύκλου circle ABC, and the straight-line AC has been drawn
τμήματι γωνίᾳ τῇ ὑπὸ ΑΒΓ. ἀλλ᾿ ἡ ὑπὸ ΘΑΓ τῇ ὑπὸ ΔΕΖ across (the circle) from the point of contact A, (angle)
ἐστιν ἴση· καὶ ἡ ὑπὸ ΑΒΓ ἄρα γωνία τῇ ὑπὸ ΔΕΖ ἐστιν HAC is thus equal to the angle ABC in the alternate
ἴση. διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΖΕ ἐστιν segment of the circle [Prop. 3.32]. But, HAC is equal to
ἴση· καὶ λοιπὴ ἄρα ἡ ὑπὸ ΒΑΓ λοιπῇ τῇ ὑπὸ ΕΔΖ ἐστιν ἴση DEF. Thus, angle ABC is also equal to DEF. So, for the
[ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ, same (reasons), ACB is also equal to DFE. Thus, the re-
καὶ ἐγγέγραπται εἰς τὸν ΑΒΓ κύκλον]. maining (angle) BAC is equal to the remaining (angle)
Εἰς τὸν δοθέντα ἄρα κύκλον τῷ δοθέντι τριγώνῳ EDF [Prop. 1.32]. [Thus, triangle ABC is equiangu-
ἰσογώνιον τρίγωνον ἐγγέγραπται· ὅπερ ἔδει ποιῆσαι. lar with triangle DEF, and has been inscribed in circle


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gþ ABC].



Thus, a triangle, equiangular with the given triangle,
has been inscribed in the given circle. (Which is) the very
thing it was required to do.

† See the footnote to Prop. 3.34.

Proposition 3
.
Περὶ τὸν δοθέντα κύκλον τῷ δοθέντι τριγώνῳ ἰσογώνιον To circumscribe a triangle, equiangular with a given
τρίγωνον περιγράψαι. triangle, about a given circle.
Θ H
Μ M
Ζ ∆ F D
Α A
Β B

Ε E
Κ K

Η G
Λ Γ Ν L C N
῎Εστω ὁ δοθεὶς κύκλος ὁ ΑΒΓ, τὸ δὲ δοθὲν τρίγωνον Let ABC be the given circle, and DEF the given tri-
τὸ ΔΕΖ· δεῖ δὴ περὶ τὸν ΑΒΓ κύκλον τῷ ΔΕΖ τριγώνῳ angle. So it is required to circumscribe a triangle, equian-
ἰσογώνιον τρίγωνον περιγράψαι. gular with triangle DEF, about circle ABC.
᾿Εκβεβλήσθω ἡ ΕΖ ἐφ᾿ ἑκάτερα τὰ μέρη κατὰ τὰ Η, Θ Let EF have been produced in each direction to
σημεῖα, καὶ εἰλήφθω τοῦ ΑΒΓ κύκλου κέντρον τὸ Κ, καὶ points G and H. And let the center K of circle ABC
διήχθω, ὡς ἔτυχεν, εὐθεῖα ἡ ΚΒ, καὶ συνεστάτω πρὸς τῇ have been found [Prop. 3.1]. And let the straight-line
ΚΒ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Κ τῇ μὲν ὑπὸ ΔΕΗ KB have been drawn, at random, across (ABC). And
γωνίᾳ ἴση ἡ ὑπὸ ΒΚΑ, τῇ δὲ ὑπὸ ΔΖΘ ἴση ἡ ὑπὸ ΒΚΓ, καὶ let (angle) BKA, equal to angle DEG, have been con-
διὰ τῶν Α, Β, Γ σημείων ἤχθωσαν ἐφαπτόμεναι τοῦ ΑΒΓ structed on the straight-line KB at the point K on it,
κύκλου αἱ ΛΑΜ, ΜΒΝ, ΝΓΛ. and (angle) BKC, equal to DFH [Prop. 1.23]. And let
Καὶ ἐπεὶ ἐφάπτονται τοῦ ΑΒΓ κύκλου αἱ ΛΜ, ΜΝ, ΝΛ the (straight-lines) LAM, MBN, and NCL have been
κατὰ τὰ Α, Β, Γ σημεῖα, ἀπὸ δὲ τοῦ Κ κέντρου ἐπὶ τὰ Α, Β, drawn through the points A, B, and C (respectively),
Γ σημεῖα ἐπεζευγμέναι εἰσὶν αἱ ΚΑ, ΚΒ, ΚΓ, ὀρθαὶ ἄρα εἰσὶν touching the circle ABC. †
αἱ πρὸς τοῖς Α, Β, Γ σημείοις γωνίαι. καὶ ἐπεὶ τοῦ ΑΜΒΚ And since LM, MN, and NL touch circle ABC at
τετραπλεύρου αἱ τέσσαρες γωνίαι τέτρασιν ὀρθαῖς ἴσαι εἰσίν, points A, B, and C (respectively), and KA, KB, and
ἐπειδήπερ καὶ εἰς δύο τρίγωνα διαιρεῖται τὸ ΑΜΒΚ, καί εἰσιν KC are joined from the center K to points A, B, and
ὀρθαὶ αἱ ὑπὸ ΚΑΜ, ΚΒΜ γωνίαι, λοιπαὶ ἄρα αἱ ὑπὸ ΑΚΒ, C (respectively), the angles at points A, B, and C are
ΑΜΒ δυσὶν ὀρθαῖς ἴσαι εἰσίν. εἰσὶ δὲ καὶ αἱ ὑπὸ ΔΕΗ, thus right-angles [Prop. 3.18]. And since the (sum of the)
ΔΕΖ δυσὶν ὀρθαῖς ἴσαι· αἱ ἄρα ὑπὸ ΑΚΒ, ΑΜΒ ταῖς ὑπὸ four angles of quadrilateral AMBK is equal to four right-
ΔΕΗ, ΔΕΖ ἴσαι εἰσίν, ὧν ἡ ὑπὸ ΑΚΒ τῇ ὑπὸ ΔΕΗ ἐστιν angles, inasmuch as AMBK (can) also (be) divided into
ἴση· λοιπὴ ἄρα ἡ ὑπὸ ΑΜΒ λοιπῇ τῇ ὑπὸ ΔΕΖ ἐστιν ἴση. two triangles [Prop. 1.32], and angles KAM and KBM
ὁμοίως δὴ δειχθήσεται, ὅτι καὶ ἡ ὑπὸ ΛΝΒ τῇ ὑπὸ ΔΖΕ are (both) right-angles, the (sum of the) remaining (an-
ἐστιν ἴση· καὶ λοιπὴ ἄρα ἡ ὑπὸ ΜΛΝ [λοιπῇ] τῇ ὑπὸ ΕΔΖ gles), AKB and AMB, is thus equal to two right-angles.
ἐστιν ἴση. ἰσογώνιον ἄρα ἐστὶ τὸ ΛΜΝ τρίγωνον τῷ ΔΕΖ And DEG and DEF is also equal to two right-angles
τριγώνῳ· καὶ περιγέγραπται περὶ τὸν ΑΒΓ κύκλον. [Prop. 1.13]. Thus, AKB and AMB is equal to DEG
Περὶ τὸν δοθέντα ἄρα κύκλον τῷ δοθέντι τριγώνῳ and DEF, of which AKB is equal to DEG. Thus, the re-
ἰσογώνιον τρίγωνον περιγέγραπται· ὅπερ ἔδει ποιῆσαι. mainder AMB is equal to the remainder DEF. So, sim-
ilarly, it can be shown that LNB is also equal to DFE.
Thus, the remaining (angle) MLN is also equal to the


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dþ [remaining] (angle) EDF [Prop. 1.32]. Thus, triangle
LMN is equiangular with triangle DEF. And it has been
drawn around circle ABC.
Thus, a triangle, equiangular with the given triangle,
has been circumscribed about the given circle. (Which is)
the very thing it was required to do.

† See the footnote to Prop. 3.34.

.
Proposition 4
Εἰς τὸ δοθὲν τρίγωνον κύκλον ἐγγράψαι. To inscribe a circle in a given triangle.
Α A



Ε E
Η G

∆ D




Β Γ B C
Ζ F
῎Εστω τὸ δοθὲν τρίγωνον τὸ ΑΒΓ· δεῖ δὴ εἰς τὸ ΑΒΓ Let ABC be the given triangle. So it is required to
τρίγωνον κύκλον ἐγγράψαι. inscribe a circle in triangle ABC.
Τετμήσθωσαν αἱ ὑπὸ ΑΒΓ, ΑΓΒ γωνίαι δίχα ταῖς ΒΔ, Let the angles ABC and ACB have been cut in half by
ΓΔ εὐθείαις, καὶ συμβαλλέτωσαν ἀλλήλαις κατὰ τὸ Δ the straight-lines BD and CD (respectively) [Prop. 1.9],
σημεῖον, καὶ ἤχθωσαν ἀπὸ τοῦ Δ ἐπὶ τὰς ΑΒ, ΒΓ, ΓΑ and let them meet one another at point D, and let DE,
εὐθείας κάθετοι αἱ ΔΕ, ΔΖ, ΔΗ. DF, and DG have been drawn from point D, perpendic-
Καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΑΒΔ γωνία τῇ ὑπὸ ΓΒΔ, ular to the straight-lines AB, BC, and CA (respectively)
ἐστὶ δὲ καὶ ὀρθὴ ἡ ὑπὸ ΒΕΔ ὀρθῇ τῇ ὑπὸ ΒΖΔ ἴση, δύο [Prop. 1.12].
δὴ τρίγωνά ἐστι τὰ ΕΒΔ, ΖΒΔ τὰς δύο γωνίας ταῖς δυσὶ And since angle ABD is equal to CBD, and the right-
γωνίαις ἴσας ἔχοντα καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην τὴν angle BED is also equal to the right-angle BFD, EBD
ὑποτείνουσαν ὑπὸ μίαν τῶν ἴσων γωνιῶν κοινὴν αὐτῶν τὴν and FBD are thus two triangles having two angles equal
ΒΔ· καὶ τὰς λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας to two angles, and one side equal to one side—the (one)
ἕξουσιν· ἴση ἄρα ἡ ΔΕ τῇ ΔΖ. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΔΗ subtending one of the equal angles (which is) common to
τῇ ΔΖ ἐστιν ἴση. αἱ τρεῖς ἄρα εὐθεῖαι αἱ ΔΕ, ΔΖ, ΔΗ the (triangles)—(namely), BD. Thus, they will also have
ἴσαι ἀλλήλαις εἰσίν· ὁ ἄρα κέντρῷ τῷ Δ καὶ διαστήματι ἑνὶ the remaining sides equal to the (corresponding) remain-
τῶν Ε, Ζ, Η κύκλος γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν ing sides [Prop. 1.26]. Thus, DE (is) equal to DF. So,
σημείων καὶ ἐφάψεται τῶν ΑΒ, ΒΓ, ΓΑ εὐθειῶν διὰ τὸ for the same (reasons), DG is also equal to DF. Thus,
ὀρθὰς εἶναι τὰς πρὸς τοῖς Ε, Ζ, Η σημείοις γωνίας. εἰ γὰρ the three straight-lines DE, DF, and DG are equal to
τεμεῖ αὐτάς, ἔσται ἡ τῇ διαμέτρῳ τοῦ κύκλου πρὸς ὀρθὰς one another. Thus, the circle drawn with center D, and
†
ἀπ᾿ ἄκρας ἀγομένη ἐντὸς πίπτουσα τοῦ κύκλου· ὅπερ ἄτο- radius one of E, F, or G, will also go through the re-
πον ἐδείχθη· οὐκ ἄρα ὁ κέντρῳ τῷ Δ διαστήματι δὲ ἑνὶ τῶν maining points, and will touch the straight-lines AB, BC,
Ε, Ζ, Η γραφόμενος κύκλος τεμεῖ τὰς ΑΒ, ΒΓ, ΓΑ εὐθείας· and CA, on account of the angles at E, F, and G being
ἐφάψεται ἄρα αὐτῶν, καὶ ἔσται ὁ κύκλος ἐγγεγραμμένος εἰς right-angles. For if it cuts (one of) them then it will be
τὸ ΑΒΓ τρίγωνον. ἐγγεγράφθω ὡς ὁ ΖΗΕ. a (straight-line) drawn at right-angles to a diameter of
Εἰς ἄρα τὸ δοθὲν τρίγωνον τὸ ΑΒΓ κύκλος ἐγγέγραπται the circle, from its extremity, falling inside the circle. The
ὁ ΕΖΗ· ὅπερ ἔδει ποιῆσαι. very thing was shown (to be) absurd [Prop. 3.16]. Thus,
the circle drawn with center D, and radius one of E, F,


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ELEMENTS BOOK 4



or G, does not cut the straight-lines AB, BC, and CA.
eþ FGE (in the figure).
Thus, it will touch them and will be the circle inscribed
in triangle ABC. Let it have been (so) inscribed, like

Thus, the circle EFG has been inscribed in the given
triangle ABC. (Which is) the very thing it was required
to do.

† Here, and in the following propositions, it is understood that the radius is actually one of DE, DF , or DG.

.
Proposition 5
Περὶ τὸ δοθὲν τρίγωνον κύκλον περιγράψαι. To circumscribe a circle about a given triangle.
Α Α Α A A A
∆ D
Β B
∆ Ε Ε D E E
∆ D
Ε E
Β Γ Γ B C C
Ζ Ζ Ζ F F F
Β B
Γ C
῎Εστω τὸ δοθὲν τρίγωνον τὸ ΑΒΓ· δεῖ δὲ περὶ τὸ δοθὲν Let ABC be the given triangle. So it is required to
τρίγωνον τὸ ΑΒΓ κύκλον περιγράψαι. circumscribe a circle about the given triangle ABC.
Τετμήσθωσαν αἱ ΑΒ, ΑΓ εὐθεῖαι δίχα κατὰ τὰ Δ, Ε Let the straight-lines AB and AC have been cut in
σημεῖα, καὶ ἀπὸ τῶν Δ, Ε σημείων ταῖς ΑΒ, ΑΓ πρὸς ὀρθὰς half at points D and E (respectively) [Prop. 1.10]. And
ἤχθωσαν αἱ ΔΖ, ΕΖ· συμπεσοῦνται δὴ ἤτοι ἐντὸς τοῦ ΑΒΓ let DF and EF have been drawn from points D and E,
τριγώνου ἢ ἐπὶ τῆς ΒΓ εὐθείας ἢ ἐκτὸς τῆς ΒΓ. at right-angles to AB and AC (respectively) [Prop. 1.11].
Συμπιπτέτωσαν πρότερον ἐντὸς κατὰ τὸ Ζ, καὶ ἐπεζεύχθ- So (DF and EF) will surely either meet inside triangle
ωσαν αἱ ΖΒ, ΖΓ, ΖΑ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΔ τῇ ΔΒ, κοινὴ ABC, on the straight-line BC, or beyond BC.
δὲ καὶ πρὸς ὀρθὰς ἡ ΔΖ, βάσις ἄρα ἡ ΑΖ βάσει τῇ ΖΒ ἐστιν Let them, first of all, meet inside (triangle ABC) at
ἴση. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἡ ΓΖ τῇ ΑΖ ἐστιν ἴση· (point) F, and let FB, FC, and FA have been joined.
ὥστε καὶ ἡ ΖΒ τῇ ΖΓ ἐστιν ἴση· αἱ τρεῖς ἄρα αἱ ΖΑ, ΖΒ, ΖΓ And since AD is equal to DB, and DF is common and
ἴσαι ἀλλήλαις εἰσίν. ὁ ἄρα κέντρῳ τῷ Ζ διαστήματι δὲ ἑνὶ at right-angles, the base AF is thus equal to the base FB
τῶν Α, Β, Γ κύκλος γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν [Prop. 1.4]. So, similarly, we can show that CF is also
σημείων, καὶ ἔσται περιγεγραμμένος ὁ κύκλος περὶ τὸ ΑΒΓ equal to AF. So that FB is also equal to FC. Thus, the
τρίγωνον. περιγεγράφθω ὡς ὁ ΑΒΓ. three (straight-lines) FA, FB, and FC are equal to one
᾿Αλλὰ δὴ αἱ ΔΖ, ΕΖ συμπιπτέτωσαν ἐπὶ τῆς ΒΓ εὐθείας another. Thus, the circle drawn with center F, and radius
κατὰ τὸ Ζ, ὡς ἔχει ἐπὶ τῆς δευτέρας καταγραφῆς, καὶ one of A, B, or C, will also go through the remaining
ἐπεζεύχθω ἡ ΑΖ. ὁμοίως δὴ δείξομεν, ὅτι τὸ Ζ σημεῖον points. And the circle will have been circumscribed about
κέντρον ἐστὶ τοῦ περὶ τὸ ΑΒΓ τρίγωνον περιγραφομένου triangle ABC. Let it have been (so) circumscribed, like
κύκλου. ABC (in the first diagram from the left).
᾿Αλλὰ δὴ αἱ ΔΖ, ΕΖ συμπιπτέτωσαν ἐκτὸς τοῦ ΑΒΓ And so, let DF and EF meet on the straight-line
τριγώνου κατὰ τὸ Ζ πάλιν, ὡς ἔχει ἐπὶ τῆς τρίτης κατα- BC at (point) F, like in the second diagram (from the
γραφῆς, καί ἐπεζεύχθωσαν αἱ ΑΖ, ΒΖ, ΓΖ. καὶ ἐπεὶ πάλιν left). And let AF have been joined. So, similarly, we can
ἴση ἐστὶν ἡ ΑΔ τῇ ΔΒ, κοινὴ δὲ καὶ πρὸς ὀρθὰς ἡ ΔΖ, βάσις show that point F is the center of the circle circumscribed
ἄρα ἡ ΑΖ βάσει τῇ ΒΖ ἐστιν ἴση. ὁμοίως δὴ δείξομεν, ὅτι about triangle ABC.
καὶ ἡ ΓΖ τῇ ΑΖ ἐστιν ἴση· ὥστε καὶ ἡ ΒΖ τῇ ΖΓ ἐστιν ἴση· And so, let DF and EF meet outside triangle ABC,
ὁ ἄρα [πάλιν] κέντρῳ τῷ Ζ διαστήματι δὲ ἑνὶ τῶν ΖΑ, ΖΒ, again at (point) F, like in the third diagram (from the
ΖΓ κύκλος γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν σημείων, left). And let AF, BF, and CF have been joined. And,
καὶ ἔσται περιγεγραμμένος περὶ τὸ ΑΒΓ τρίγωνον. again, since AD is equal to DB, and DF is common and
Περὶ τὸ δοθὲν ἄρα τρίγωνον κύκλος περιγέγραπται· at right-angles, the base AF is thus equal to the base BF
ὅπερ ἔδει ποιῆσαι. [Prop. 1.4]. So, similarly, we can show that CF is also
equal to AF. So that BF is also equal to FC. Thus,

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þ [again] the circle drawn with center F, and radius one



of FA, FB, and FC, will also go through the remaining
points. And it will have been circumscribed about trian-
gle ABC.
Thus, a circle has been circumscribed about the given
triangle. (Which is) the very thing it was required to do.

.
Proposition 6
Εἰς τὸν δοθέντα κύκλον τετράγωνον ἐγγράψαι. To inscribe a square in a given circle.
Α A






Ε E
Β ∆ B D







Γ C
῎Εστω ἡ δοθεὶς κύκλος ὁ ΑΒΓΔ· δεῖ δὴ εἰς τὸν ΑΒΓΔ Let ABCD be the given circle. So it is required to
κύκλον τετράγωνον ἐγγράψαι. inscribe a square in circle ABCD.
῎Ηχθωσαν τοῦ ΑΒΓΔ κύκλου δύο διάμετροι πρὸς ὀρθὰς Let two diameters of circle ABCD, AC and BD, have
†
ἀλλήλαις αἱ ΑΓ, ΒΔ, καὶ ἐπεζεύχθωσαν αἱ ΑΒ, ΒΓ, ΓΔ, been drawn at right-angles to one another. And let AB,
ΔΑ. BC, CD, and DA have been joined.
Καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΕ τῇ ΕΔ· κέντρον γὰρ τὸ Ε· κοινὴ And since BE is equal to ED, for E (is) the center
δὲ καὶ πρὸς ὀρθὰς ἡ ΕΑ, βάσις ἄρα ἡ ΑΒ βάσει τῇ ΑΔ ἴση (of the circle), and EA is common and at right-angles,
ἐστίν. διὰ τὰ αὐτὰ δὴ καὶ ἑκατέρα τῶν ΒΓ, ΓΔ ἑκατέρᾳ the base AB is thus equal to the base AD [Prop. 1.4].
τῶν ΑΒ, ΑΔ ἴση ἐστίν· ἰσόπλευρον ἄρα ἐστὶ τὸ ΑΒΓΔ So, for the same (reasons), each of BC and CD is equal
τετράπλευρον. λέγω δή, ὅτι καὶ ὀρθογώνιον. ἐπεὶ γὰρ ἡ to each of AB and AD. Thus, the quadrilateral ABCD
ΒΔ εὐθεῖα διάμετρός ἐστι τοῦ ΑΒΓΔ κύκλου, ἡμικύκλιον is equilateral. So I say that (it is) also right-angled. For
ἄρα ἐστὶ τὸ ΒΑΔ· ὀρθὴ ἄρα ἡ ὑπὸ ΒΑΔ γωνία. διὰ τὰ since the straight-line BD is a diameter of circle ABCD,
αὐτὰ δὴ καὶ ἑκάστη τῶν ὑπὸ ΑΒΓ, ΒΓΔ, ΓΔΑ ὀρθή ἐστιν· BAD is thus a semi-circle. Thus, angle BAD (is) a right-
ὀρθογώνιον ἄρα ἐστὶ τὸ ΑΒΓΔ τετράπλευρον. ἐδείχθη δὲ angle [Prop. 3.31]. So, for the same (reasons), (angles)
καὶ ἰσόπλευρον· τετράγωνον ἄρα ἐστίν. καὶ ἐγγέγραπται εἰς ABC, BCD, and CDA are also each right-angles. Thus,
τὸν ΑΒΓΔ κύκλον. the quadrilateral ABCD is right-angled. And it was also
zþ given circle. (Which is) the very thing it was required
Εἰς ἄρα τὸν δοθέντα κύκλον τετράγωνον ἐγγέγραπται shown (to be) equilateral. Thus, it is a square [Def. 1.22].
τὸ ΑΒΓΔ· ὅπερ ἔδει ποιῆσαι. And it has been inscribed in circle ABCD.
Thus, the square ABCD has been inscribed in the
to do.

† Presumably, by finding the center of the circle [Prop. 3.1], drawing a line through it, and then drawing a second line through it, at right-angles
to the first [Prop. 1.11].

Proposition 7
.
Περὶ τὸν δοθέντα κύκλον τετράγωνον περιγράψαι. To circumscribe a square about a given circle.


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῎Εστω ὁ δοθεὶς κύκλος ὁ ΑΒΓΔ· δεῖ δὴ περὶ τὸν ΑΒΓΔ Let ABCD be the given circle. So it is required to
κύκλον τετράγωνον περιγράψαι. circumscribe a square about circle ABCD.
Η Α Ζ G A F






Ε E
Β ∆ B D







Θ Γ Κ H C K
῎Ηχθωσαν τοῦ ΑΒΓΔ κύκλου δύο διάμετροι πρὸς ὀρθὰς Let two diameters of circle ABCD, AC and BD, have
†
ἀλλήλαις αἱ ΑΓ, ΒΔ, καὶ διὰ τῶν Α, Β, Γ, Δ σημείων ἤχθω- been drawn at right-angles to one another. And let FG,
σαν ἐφαπτόμεναι τοῦ ΑΒΓΔ κύκλου αἱ ΖΗ, ΗΘ, ΘΚ, ΚΖ. GH, HK, and KF have been drawn through points A,
᾿Επεὶ οὖν ἐφάπτεται ἡ ΖΗ τοῦ ΑΒΓΔ κύκλου, ἀπὸ δὲ B, C, and D (respectively), touching circle ABCD. ‡
τοῦ Ε κέντρου ἐπὶ τὴν κατὰ τὸ Α ἐπαφὴν ἐπέζευκται ἡ Therefore, since FG touches circle ABCD, and EA
ΕΑ, αἱ ἄρα πρὸς τῷ Α γωνίαι ὀρθαί εἰσιν. διὰ τὰ αὐτὰ has been joined from the center E to the point of contact
δὴ καὶ αἱ πρὸς τοῖς Β, Γ, Δ σημείοις γωνίαι ὀρθαί εἰσιν. A, the angles at A are thus right-angles [Prop. 3.18]. So,
καὶ ἐπεὶ ὀρθή ἐστιν ἡ ὑπὸ ΑΕΒ γωνία, ἐστὶ δὲ ὀρθὴ καὶ ἡ for the same (reasons), the angles at points B, C, and
ὑπὸ ΕΒΗ, παράλληλος ἄρα ἐστὶν ἡ ΗΘ τῇ ΑΓ. διὰ τὰ αὐτὰ D are also right-angles. And since angle AEB is a right-
δὴ καὶ ἡ ΑΓ τῇ ΖΚ ἐστι παράλληλος. ὥστε καὶ ἡ ΗΘ τῇ angle, and EBG is also a right-angle, GH is thus parallel
ΖΚ ἐστι παράλληλος. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἑκατέρα to AC [Prop. 1.29]. So, for the same (reasons), AC is
τῶν ΗΖ, ΘΚ τῇ ΒΕΔ ἐστι παράλληλος. παραλληλόγραμμα also parallel to FK. So that GH is also parallel to FK
ἄρα ἐστὶ τὰ ΗΚ, ΗΓ, ΑΚ, ΖΒ, ΒΚ· ἴση ἄρα ἐστὶν ἡ μὲν [Prop. 1.30]. So, similarly, we can show that GF and
ΗΖ τῇ ΘΚ, ἡ δὲ ΗΘ τῇ ΖΚ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΓ τῇ HK are each parallel to BED. Thus, GK, GC, AK, FB,
ΒΔ, ἀλλὰ καὶ ἡ μὲν ΑΓ ἑκατέρᾳ τῶν ΗΘ, ΖΚ, ἡ δὲ ΒΔ and BK are (all) parallelograms. Thus, GF is equal to
ἑκατέρᾳ τῶν ΗΖ, ΘΚ ἐστιν ἴση [καὶ ἑκατέρα ἄρα τῶν ΗΘ, HK, and GH to FK [Prop. 1.34]. And since AC is equal
ΖΚ ἑκατέρᾳ τῶν ΗΖ, ΘΚ ἐστιν ἴση], ἰσόπλευρον ἄρα ἐστὶ τὸ to BD, but AC (is) also (equal) to each of GH and FK,
ΖΗΘΚ τετράπλευρον. λέγω δή, ὅτι καὶ ὀρθογώνιον. ἐπεὶ and BD is equal to each of GF and HK [Prop. 1.34]
γὰρ παραλληλόγραμμόν ἐστι τὸ ΗΒΕΑ, καί ἐστιν ὀρθὴ ἡ [and each of GH and FK is thus equal to each of GF
ὑπὸ ΑΕΒ, ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΑΗΒ. ὁμοίως δὴ δείξομεν, and HK], the quadrilateral FGHK is thus equilateral.
ὅτι καὶ αἱ πρὸς τοῖς Θ, Κ, Ζ γωνίαι ὀρθαί εἰσιν. ὀρθογώνιον So I say that (it is) also right-angled. For since GBEA
ἄρα ἐστὶ τὸ ΖΗΘΚ. ἐδείχθη δὲ καὶ ἰσόπλευρον· τετράγωνον is a parallelogram, and AEB is a right-angle, AGB is
ἄρα ἐστίν. καὶ περιγέγραπται περὶ τὸν ΑΒΓΔ κύκλον. thus also a right-angle [Prop. 1.34]. So, similarly, we can
Περὶ τὸν δοθέντα ἄρα κύκλον τετράγωνον περιγέγραπται· show that the angles at H, K, and F are also right-angles.
ὅπερ ἔδει ποιῆσαι. Thus, FGHK is right-angled. And it was also shown (to
be) equilateral. Thus, it is a square [Def. 1.22]. And it
has been circumscribed about circle ABCD.
Thus, a square has been circumscribed about the
given circle. (Which is) the very thing it was required
to do.

† See the footnote to the previous proposition.
‡ See the footnote to Prop. 3.34.





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.
Proposition 8
Εἰς τὸ δοθὲν τετράγωνον κύκλον ἐγγράψαι. To inscribe a circle in a given square.
῎Εστω τὸ δοθὲν τετράγωνον τὸ ΑΒΓΔ. δεῖ δὴ εἰς τὸ Let the given square be ABCD. So it is required to
ΑΒΓΔ τετράγωνον κύκλον ἐγγράψαι. inscribe a circle in square ABCD.
Α Ε ∆ A E D






Η G
Ζ Κ F K







Β Θ Γ B H C
Τετμήσθω ἑκατέρα τῶν ΑΔ, ΑΒ δίχα κατὰ τὰ Ε, Ζ Let AD and AB each have been cut in half at points E
σημεῖα, καὶ διὰ μὲν τοῦ Ε ὁποτέρᾳ τῶν ΑΒ, ΓΔ παράλληλος and F (respectively) [Prop. 1.10]. And let EH have been
ἤχθω ὁ ΕΘ, διὰ δὲ τοῦ Ζ ὁποτέρᾳ τῶν ΑΔ, ΒΓ παράλληλος drawn through E, parallel to either of AB or CD, and let
ἤχθω ἡ ΖΚ· παραλληλόγραμμον ἄρα ἐστὶν ἕκαστον τῶν FK have been drawn through F, parallel to either of AD
ΑΚ, ΚΒ, ΑΘ, ΘΔ, ΑΗ, ΗΓ, ΒΗ, ΗΔ, καὶ αἱ ἀπεναντίον or BC [Prop. 1.31]. Thus, AK, KB, AH, HD, AG, GC,
αὐτῶν πλευραὶ δηλονότι ἴσαι [εἰσίν]. καὶ ἐπεὶ ἴση ἐστὶν ἡ BG, and GD are each parallelograms, and their opposite
ΑΔ τῇ ΑΒ, καί ἐστι τῆς μὲν ΑΔ ἡμίσεια ἡ ΑΕ, τῆς δὲ ΑΒ sides [are] manifestly equal [Prop. 1.34]. And since AD
ἡμίσεια ἡ ΑΖ, ἴση ἄρα καὶ ἡ ΑΕ τῇ ΑΖ· ὥστε καὶ αἱ ἀπε- is equal to AB, and AE is half of AD, and AF half of
ναντίον· ἴση ἄρα καὶ ἡ ΖΗ τῇ ΗΕ. ὁμοίως δὴ δείξομεν, ὅτι AB, AE (is) thus also equal to AF. So that the opposite
καὶ ἑκατέρα τῶν ΗΘ, ΗΚ ἑκατέρᾳ τῶν ΖΗ, ΗΕ ἐστιν ἴση· (sides are) also (equal). Thus, FG (is) also equal to GE.
αἱ τέσσαρες ἄρα αἱ ΗΕ, ΗΖ, ΗΘ, ΗΚ ἴσαι ἀλλήλαις [εἰσίν]. So, similarly, we can also show that each of GH and GK
ὁ ἄρα κέντρῳ μὲν τῷ Η διαστήματι δὲ ἑνὶ τῶν Ε, Ζ, Θ, Κ is equal to each of FG and GE. Thus, the four (straight-
κύκλος γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν σημείων· καὶ lines) GE, GF, GH, and GK [are] equal to one another.
ἐφάψεται τῶν ΑΒ, ΒΓ, ΓΔ, ΔΑ εὐθειῶν διὰ τὸ ὀρθὰς εἶναι Thus, the circle drawn with center G, and radius one of
τὰς πρὸς τοῖς Ε, Ζ, Θ, Κ γωνίας· εἰ γὰρ τεμεῖ ὁ κύκλος τὰς E, F, H, or K, will also go through the remaining points.
ΑΒ, ΒΓ, ΓΔ, ΔΑ, ἡ τῇ διαμέτρῳ τοῦ κύκλου πρὸς ὀρθὰς And it will touch the straight-lines AB, BC, CD, and
ἀπ᾿ ἄκρας ἀγομένη ἐντὸς πεσεῖται τοῦ κύκλου· ὅπερ ἄτοπον DA, on account of the angles at E, F, H, and K being
ἐδείχθη. οὐκ ἄρα ὁ κέντρῳ τῷ Η διαστήματι δὲ ἑνὶ τῶν Ε, right-angles. For if the circle cuts AB, BC, CD, or DA,
Ζ, Θ, Κ κύκλος γραφόμενος τεμεῖ τὰς ΑΒ, ΒΓ, ΓΔ, ΔΑ then a (straight-line) drawn at right-angles to a diameter
jþ CD, or DA. Thus, it will touch them, and will have been
εὐθείας. ἐφάψεται ἄρα αὐτῶν καὶ ἔσται ἐγγεγραμμένος εἰς of the circle, from its extremity, will fall inside the circle.
τὸ ΑΒΓΔ τετράγωνον. The very thing was shown (to be) absurd [Prop. 3.16].
Εἰς ἄρα τὸ δοθὲν τετράγωνον κύκλος ἐγγέγραπται· Thus, the circle drawn with center G, and radius one of
ὅπερ ἔδει ποιῆσαι. E, F, H, or K, does not cut the straight-lines AB, BC,

inscribed in the square ABCD.
Thus, a circle has been inscribed in the given square.
(Which is) the very thing it was required to do.

Proposition 9
.
Περὶ τὸ δοθὲν τετράγωνον κύκλον περιγράψαι. To circumscribe a circle about a given square.
῎Εστω τὸ δοθὲν τετράγωνον τὸ ΑΒΓΔ· δεῖ δὴ περὶ τὸ Let ABCD be the given square. So it is required to
ΑΒΓΔ τετράγωνον κύκλον περιγράψαι. circumscribe a circle about square ABCD.


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᾿Επιζευχθεῖσαι γὰρ αἱ ΑΓ, ΒΔ τεμνέτωσαν ἀλλήλας AC and BD being joined, let them cut one another at
κατὰ τὸ Ε. E.
Α A






Ε E
Β ∆ B D







Γ C
Καὶ ἐπεὶ ἴση ἐστὶν ἡ ΔΑ τῇ ΑΒ, κοινὴ δὲ ἡ ΑΓ, δύο And since DA is equal to AB, and AC (is) common,
δὴ αἱ ΔΑ, ΑΓ δυσὶ ταῖς ΒΑ, ΑΓ ἴσαι εἰσίν· καὶ βάσις ἡ the two (straight-lines) DA, AC are thus equal to the two
ΔΓ βάσει τῇ ΒΓ ἴση· γωνία ἄρα ἡ ὑπὸ ΔΑΓ γωνίᾳ τῇ ὑπὸ (straight-lines) BA, AC. And the base DC (is) equal to
ΒΑΓ ἴση ἐστίν· ἡ ἄρα ὑπὸ ΔΑΒ γωνία δίχα τέτμηται ὑπὸ the base BC. Thus, angle DAC is equal to angle BAC
τῆς ΑΓ. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἑκάστη τῶν ὑπὸ ΑΒΓ, [Prop. 1.8]. Thus, the angle DAB has been cut in half
ΒΓΔ, ΓΔΑ δίχα τέτμηται ὑπὸ τῶν ΑΓ, ΔΒ εὐθειῶν. καὶ by AC. So, similarly, we can show that ABC, BCD, and
ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΔΑΒ γωνία τῇ ὑπὸ ΑΒΓ, καί ἐστι CDA have each been cut in half by the straight-lines AC
τῆς μὲν ὑπὸ ΔΑΒ ἡμίσεια ἡ ὑπὸ ΕΑΒ, τῆς δὲ ὑπὸ ΑΒΓ and DB. And since angle DAB is equal to ABC, and
ἡμίσεια ἡ ὑπὸ ΕΒΑ, καὶ ἡ ὑπὸ ΕΑΒ ἄρα τῇ ὑπὸ ΕΒΑ ἐστιν EAB is half of DAB, and EBA half of ABC, EAB is
ἴση· ὥστε καὶ πλευρὰ ἡ ΕΑ τῇ ΕΒ ἐστιν ἴση. ὁμοίως δὴ thus also equal to EBA. So that side EA is also equal
δείξομεν, ὅτι καὶ ἑκατέρα τῶν ΕΑ, ΕΒ [εὐθειῶν] ἑκατέρᾳ to EB [Prop. 1.6]. So, similarly, we can show that each
τῶν ΕΓ, ΕΔ ἴση ἐστίν. αἱ τέσσαρες ἄρα αἱ ΕΑ, ΕΒ, ΕΓ, of the [straight-lines] EA and EB are also equal to each
iþ
ΕΔ ἴσαι ἀλλήλαις εἰσίν. ὁ ἄρα κέντρῳ τῷ Ε καὶ διαστήματι of EC and ED. Thus, the four (straight-lines) EA, EB,
ἑνὶ τῶν Α, Β, Γ, Δ κύκλος γραφόμενος ἥξει καὶ διὰ τῶν EC, and ED are equal to one another. Thus, the circle
λοιπῶν σημείων καὶ ἔσται περιγεγραμμένος περὶ τὸ ΑΒΓΔ drawn with center E, and radius one of A, B, C, or D,
τετράγωνον. περιγεγράφθω ὡς ὁ ΑΒΓΔ. will also go through the remaining points, and will have
Περὶ τὸ δοθὲν ἄρα τετράγωνον κύκλος περιγέγραπται· been circumscribed about the square ABCD. Let it have
ὅπερ ἔδει ποιῆσαι. been (so) circumscribed, like ABCD (in the figure).
Thus, a circle has been circumscribed about the given
square. (Which is) the very thing it was required to do.

.
Proposition 10
᾿Ισοσκελὲς τρίγωνον συστήσασθαι ἔχον ἑκατέραν τῶν To construct an isosceles triangle having each of the
πρὸς τῇ βάσει γωνιῶν διπλασίονα τῆς λοιπῆς. angles at the base double the remaining (angle).
᾿Εκκείσθω τις εὐθεῖα ἡ ΑΒ, καὶ τετμήσθω κατὰ τὸ Let some straight-line AB be taken, and let it have
Γ σημεῖον, ὥστε τὸ ὑπὸ τῶν ΑΒ, ΒΓ περιεχόμενον been cut at point C so that the rectangle contained by
ὀρθογώνιον ἴσον εἶναι τῷ ἀπὸ τῆς ΓΑ τετραγώνῳ· καὶ AB and BC is equal to the square on CA [Prop. 2.11].
κέντρῳ τῷ Α καὶ διαστήματι τῷ ΑΒ κύκλος γεγράφθω And let the circle BDE have been drawn with center A,
ὁ ΒΔΕ, καὶ ἐνηρμόσθω εἰς τὸν ΒΔΕ κύκλον τῇ ΑΓ εὐθείᾳ and radius AB. And let the straight-line BD, equal to
μὴ μείζονι οὔσῃ τῆς τοῦ ΒΔΕ κύκλου διαμέτρου ἴση εὐθεῖα the straight-line AC, being not greater than the diame-
ἡ ΒΔ· καὶ ἐπεζεύχθωσαν αἱ ΑΔ, ΔΓ, καὶ περιγεγράφθω ter of circle BDE, have been inserted into circle BDE
περὶ τὸ ΑΓΔ τρίγωνον κύκλος ὁ ΑΓΔ. [Prop. 4.1]. And let AD and DC have been joined. And
let the circle ACD have been circumscribed about trian-
gle ACD [Prop. 4.5].



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Β ELEMENTS BOOK 4
B
Γ C
∆ D

Α A


Ε E






Καὶ ἐπεὶ τὸ ὑπὸ τῶν ΑΒ, ΒΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς And since the (rectangle contained) by AB and BC
ΑΓ, ἴση δὲ ἡ ΑΓ τῇ ΒΔ, τὸ ἄρα ὑπὸ τῶν ΑΒ, ΒΓ ἴσον is equal to the (square) on AC, and AC (is) equal to
ἐστὶ τῷ ἀπὸ τῆς ΒΔ. καὶ ἐπεὶ κύκλου τοῦ ΑΓΔ εἴληπταί τι BD, the (rectangle contained) by AB and BC is thus
σημεῖον ἐκτὸς τὸ Β, καὶ ἀπὸ τοῦ Β πρὸς τὸν ΑΓΔ κύκλον equal to the (square) on BD. And since some point B
προσπεπτώκασι δύο εὐθεῖαι αἱ ΒΑ, ΒΔ, καὶ ἡ μὲν αὐτῶν has been taken outside of circle ACD, and two straight-
τέμνει, ἡ δὲ προσπίπτει, καί ἐστι τὸ ὑπὸ τῶν ΑΒ, ΒΓ ἴσον lines BA and BD have radiated from B towards the cir-
τῷ ἀπὸ τῆς ΒΔ, ἡ ΒΔ ἄρα ἐφάπτεται τοῦ ΑΓΔ κύκλου. ἐπεὶ cle ACD, and (one) of them cuts (the circle), and (the
οὖν ἐφάπτεται μὲν ἡ ΒΔ, ἀπὸ δὲ τῆς κατὰ τὸ Δ ἐπαφῆς other) meets (the circle), and the (rectangle contained)
διῆκται ἡ ΔΓ, ἡ ἄρα ὑπὸ ΒΔΓ γωνιά ἴση ἐστὶ τῇ ἐν τῷ by AB and BC is equal to the (square) on BD, BD thus
ἐναλλὰξ τοῦ κύκλου τμήματι γωνίᾳ τῇ ὑπὸ ΔΑΓ. ἐπεὶ οὖν touches circle ACD [Prop. 3.37]. Therefore, since BD
ἴση ἐστὶν ἡ ὑπὸ ΒΔΓ τῇ ὑπὸ ΔΑΓ, κοινὴ προσκείσθω ἡ touches (the circle), and DC has been drawn across (the
ὑπὸ ΓΔΑ· ὅλη ἄρα ἡ ὑπὸ ΒΔΑ ἴση ἐστὶ δυσὶ ταῖς ὑπὸ ΓΔΑ, circle) from the point of contact D, the angle BDC is
ΔΑΓ. ἀλλὰ ταῖς ὑπὸ ΓΔΑ, ΔΑΓ ἴση ἐστὶν ἡ ἐκτὸς ἡ ὑπὸ thus equal to the angle DAC in the alternate segment of
ΒΓΔ· καὶ ἡ ὑπὸ ΒΔΑ ἄρα ἴση ἐστὶ τῇ ὑπὸ ΒΓΔ. ἀλλὰ ἡ the circle [Prop. 3.32]. Therefore, since BDC is equal
ὑπὸ ΒΔΑ τῇ ὑπὸ ΓΒΔ ἐστιν ἴση, ἐπεὶ καὶ πλευρὰ ἡ ΑΔ to DAC, let CDA have been added to both. Thus, the
τῇ ΑΒ ἐστιν ἴση· ὥστε καὶ ἡ ὑπὸ ΔΒΑ τῇ ὑπὸ ΒΓΔ ἐστιν whole of BDA is equal to the two (angles) CDA and
ἴση. αἱ τρεῖς ἄρα αἱ ὑπὸ ΒΔΑ, ΔΒΑ, ΒΓΑ ἴσαι ἀλλήλαις DAC. But, the external (angle) BCD is equal to CDA
εἰσίν. καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΔΒΓ γωνία τῇ ὑπὸ ΒΓΔ, and DAC [Prop. 1.32]. Thus, BDA is also equal to
ἴση ἐστὶ καὶ πλευρὰ ἡ ΒΔ πλευρᾷ τῇ ΔΓ. ἀλλὰ ἡ ΒΔ τῇ BCD. But, BDA is equal to CBD, since the side AD is
ΓΑ ὑπόκειται ἴση· καὶ ἡ ΓΑ ἄρα τῇ ΓΔ ἐστιν ἴση· ὥστε καὶ also equal to AB [Prop. 1.5]. So that DBA is also equal
γωνία ἡ ὑπὸ ΓΔΑ γωνίᾳ τῇ ὑπὸ ΔΑΓ ἐστιν ἴση· αἱ ἄρα to BCD. Thus, the three (angles) BDA, DBA, and BCD
ὑπὸ ΓΔΑ, ΔΑΓ τῆς ὑπὸ ΔΑΓ εἰσι διπλασίους. ἴση δὲ ἡ are equal to one another. And since angle DBC is equal
ὑπὸ ΒΓΔ ταῖς ὑπὸ ΓΔΑ, ΔΑΓ· καὶ ἡ ὑπὸ ΒΓΔ ἄρα τῆς ὑπὸ to BCD, side BD is also equal to side DC [Prop. 1.6].
ΓΑΔ ἐστι διπλῆ. ἴση δὲ ἡ ὑπὸ ΒΓΔ ἑκατέρᾳ τῶν ὑπὸ ΒΔΑ, But, BD was assumed (to be) equal to CA. Thus, CA
ΔΒΑ· καὶ ἑκατέρα ἄρα τῶν ὑπὸ ΒΔΑ, ΔΒΑ τῆς ὑπὸ ΔΑΒ is also equal to CD. So that angle CDA is also equal to
ἐστι διπλῆ. angle DAC [Prop. 1.5]. Thus, CDA and DAC is double
iaþ structed having each of the angles at the base BD double
᾿Ισοσκελὲς ἄρα τρίγωνον συνέσταται τὸ ΑΒΔ ἔχον DAC. But BCD (is) equal to CDA and DAC. Thus,
ἑκατέραν τῶν πρὸς τῇ ΔΒ βάσει γωνιῶν διπλασίονα τῆς BCD is also double CAD. And BCD (is) equal to to
λοιπῆς· ὅπερ ἔδει ποιῆσαι. each of BDA and DBA. Thus, BDA and DBA are each
double DAB.
Thus, the isosceles triangle ABD has been con-

the remaining (angle). (Which is) the very thing it was
required to do.

.
Proposition 11
Εἰς τὸν δοθέντα κύκλον πεντάγωνον ἰσόπλευρόν τε καὶ To inscribe an equilateral and equiangular pentagon


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ἰσογώνιον ἐγγράψαι. in a given circle. ELEMENTS BOOK 4
Α A
Ζ F



Β Ε B E







Γ ∆ Η Θ C D G H
῎Εστω ὁ δοθεὶς κύκλος ὁ ΑΒΓΔΕ· δεῖ δὴ εἰς τὸν Let ABCDE be the given circle. So it is required to
ΑΒΓΔΕ κύκλον πεντάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον inscribed an equilateral and equiangular pentagon in cir-
ἐγγράψαι. cle ABCDE.
᾿Εκκείσθω τρίγωνον ἰσοσκελὲς τὸ ΖΗΘ διπλασίονα Let the the isosceles triangle FGH be set up hav-
ἔχον ἑκατέραν τῶν πρὸς τοῖς Η, Θ γωνιῶν τῆς πρὸς τῷ Ζ, ing each of the angles at G and H double the (angle)
καὶ ἐγγεγράφθω εἰς τὸν ΑΒΓΔΕ κύκλον τῷ ΖΗΘ τριγώνῳ at F [Prop. 4.10]. And let triangle ACD, equiangular
ἰσογώνιον τρίγωνον τὸ ΑΓΔ, ὥστε τῇ μὲν πρὸς τῷ Ζ γωνίᾳ to FGH, have been inscribed in circle ABCDE, such
ἴσην εἶναι τὴν ὑπὸ ΓΑΔ, ἑκατέραν δὲ τῶν πρὸς τοῖς Η, Θ that CAD is equal to the angle at F, and the (angles)
ἴσην ἑκατέρᾳ τῶν ὑπὸ ΑΓΔ, ΓΔΑ· καὶ ἑκατέρα ἄρα τῶν at G and H (are) equal to ACD and CDA, respectively
ὑπὸ ΑΓΔ, ΓΔΑ τῆς ὑπὸ ΓΑΔ ἐστι διπλῆ. τετμήσθω δὴ [Prop. 4.2]. Thus, ACD and CDA are each double
ἑκατέρα τῶν ὑπὸ ΑΓΔ, ΓΔΑ δίχα ὑπὸ ἑκατέρας τῶν ΓΕ, CAD. So let ACD and CDA have been cut in half by
ΔΒ εὐθειῶν, καὶ ἐπεζεύχθωσαν αἱ ΑΒ, ΒΓ, ΔΕ, ΕΑ. the straight-lines CE and DB, respectively [Prop. 1.9].
᾿Επεὶ οὖν ἑκατέρα τῶν ὑπὸ ΑΓΔ, ΓΔΑ γωνιῶν δι- And let AB, BC, DE and EA have been joined.
πλασίων ἐστὶ τῆς ὑπὸ ΓΑΔ, καὶ τετμημέναι εἰσὶ δίχα ὑπὸ Therefore, since angles ACD and CDA are each dou-
τῶν ΓΕ, ΔΒ εὐθειῶν, αἱ πέντε ἄρα γωνίαι αἱ ὑπὸ ΔΑΓ, ble CAD, and are cut in half by the straight-lines CE and
ΑΓΕ, ΕΓΔ, ΓΔΒ, ΒΔΑ ἴσαι ἀλλήλαις εἰσίν. αἱ δὲ ἴσαι DB, the five angles DAC, ACE, ECD, CDB, and BDA
γωνίαι ἐπὶ ἴσων περιφερειῶν βεβήκασιν· αἱ πέντε ἄρα πε- are thus equal to one another. And equal angles stand
ριφέρειαι αἱ ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΑ ἴσαι ἀλλήλαις εἰσίν. ὑπὸ upon equal circumferences [Prop. 3.26]. Thus, the five
δὲ τὰς ἴσας περιφερείας ἴσαι εὐθεῖαι ὑποτείνουσιν· αἱ πέντε circumferences AB, BC, CD, DE, and EA are equal to
ἄρα εὐθεῖαι αἱ ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΑ ἴσαι ἀλλήλαις εἰσίν· one another [Prop. 3.29]. Thus, the pentagon ABCDE
ἰσόπλευρον ἄρα ἐστὶ τὸ ΑΒΓΔΕ πεντάγωνον. λέγω δή, is equilateral. So I say that (it is) also equiangular. For
ὅτι καὶ ἰσογώνιον. ἐπεὶ γὰρ ἡ ΑΒ περιφέρεια τῇ ΔΕ πε- since the circumference AB is equal to the circumfer-
ριφερείᾳ ἐστὶν ἴση, κοινὴ προσκείσθω ἡ ΒΓΔ· ὅλη ἄρα ἡ ence DE, let BCD have been added to both. Thus, the
ΑΒΓΔ περιφέρια ὅλῃ τῇ ΕΔΓΒ περιφερείᾳ ἐστὶν ἴση. καὶ whole circumference ABCD is equal to the whole cir-
βέβηκεν ἐπὶ μὲν τῆς ΑΒΓΔ περιφερείας γωνία ἡ ὑπὸ ΑΕΔ, cumference EDCB. And the angle AED stands upon
ἐπὶ δὲ τῆς ΕΔΓΒ περιφερείας γωνία ἡ ὑπὸ ΒΑΕ· καὶ ἡ ὑπὸ circumference ABCD, and angle BAE upon circumfer-
ibþ
ΒΑΕ ἄρα γωνία τῇ ὑπὸ ΑΕΔ ἐστιν ἴση. διὰ τὰ αὐτὰ δὴ ence EDCB. Thus, angle BAE is also equal to AED
καὶ ἑκάστη τῶν ὑπὸ ΑΒΓ, ΒΓΔ, ΓΔΕ γωνιῶν ἑκατέρᾳ τῶν [Prop. 3.27]. So, for the same (reasons), each of the an-
ὑπὸ ΒΑΕ, ΑΕΔ ἐστιν ἴση· ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓΔΕ gles ABC, BCD, and CDE is also equal to each of BAE
πεντάγωνον. ἐδείχθη δὲ καὶ ἰσόπλευρον. and AED. Thus, pentagon ABCDE is equiangular. And
Εἰς ἄρα τὸν δοθέντα κύκλον πεντάγωνον ἰσόπλευρόν τε it was also shown (to be) equilateral.
καὶ ἰσογώνιον ἐγγέγραπται· ὅπερ ἔδει ποιῆσαι. been inscribed in the given circle. (Which is) the very
Thus, an equilateral and equiangular pentagon has
thing it was required to do.

Proposition 12
.
Περὶ τὸν δοθέντα κύκλον πεντάγωνον ἰσόπλευρόν τε To circumscribe an equilateral and equiangular pen-
καὶ ἰσογώνιον περιγράψαι. tagon about a given circle.


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Η G ELEMENTS BOOK 4

Α Ε A E


Θ Μ H M
Ζ F

Β ∆ B D



Κ Γ Λ K C L
῎Εστω ὁ δοθεὶς κύκλος ὁ ΑΒΓΔΕ· δεῖ δὲ περὶ τὸν Let ABCDE be the given circle. So it is required
ΑΒΓΔΕ κύκλον πεντάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον to circumscribe an equilateral and equiangular pentagon
περιγράψαι. about circle ABCDE.
Νενοήσθω τοῦ ἐγγεγραμμένου πενταγώνου τῶν γωνιῶν Let A, B, C, D, and E have been conceived as the an-
σημεῖα τὰ Α, Β, Γ, Δ, Ε, ὥστε ἴσας εἶναι τὰς ΑΒ, ΒΓ, gular points of a pentagon having been inscribed (in cir-
ΓΔ, ΔΕ, ΕΑ περιφερείας· καὶ διὰ τῶν Α, Β, Γ, Δ, Ε cle ABCDE) [Prop. 3.11], such that the circumferences
ἤχθωσαν τοῦ κύκλου ἐφαπτόμεναι αἱ ΗΘ, ΘΚ, ΚΛ, ΛΜ, AB, BC, CD, DE, and EA are equal. And let GH, HK,
ΜΗ, καὶ εἰλήφθω τοῦ ΑΒΓΔΕ κύκλου κέντρον τὸ Ζ, καὶ KL, LM, and MG have been drawn through (points) A,
†
ἐπεζεύχθωσαν αἱ ΖΒ, ΖΚ, ΖΓ, ΖΛ, ΖΔ. B, C, D, and E (respectively), touching the circle. And
Καὶ ἐπεὶ ἡ μὲν ΚΛ εὐθεῖα ἐφάπτεται τοῦ ΑΒΓΔΕ κατὰ let the center F of the circle ABCDE have been found
τὸ Γ, ἀπὸ δὲ τοῦ Ζ κέντρου ἐπὶ τὴν κατὰ τὸ Γ ἐπαφὴν [Prop. 3.1]. And let FB, FK, FC, FL, and FD have
ἐπέζευκται ἡ ΖΓ, ἡ ΖΓ ἄρα κάθετός ἐστιν ἐπὶ τὴν ΚΛ· ὀρθὴ been joined.
ἄρα ἐστὶν ἑκατέρα τῶν πρὸς τῷ Γ γωνιῶν. διὰ τὰ αὐτὰ δὴ And since the straight-line KL touches (circle) ABCDE
καὶ αἱ πρὸς τοῖς Β, Δ σημείοις γωνίαι ὀρθαί εἰσιν. καὶ ἐπεὶ at C, and FC has been joined from the center F to the
ὀρθή ἐστιν ἡ ὑπὸ ΖΓΚ γωνία, τὸ ἄρα ἀπὸ τῆς ΖΚ ἴσον ἐστὶ point of contact C, FC is thus perpendicular to KL
τοῖς ἀπὸ τῶν ΖΓ, ΓΚ. διὰ τὰ αὐτὰ δὴ καὶ τοῖς ἀπὸ τῶν ΖΒ, [Prop. 3.18]. Thus, each of the angles at C is a right-
ΒΚ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΖΚ· ὥστε τὰ ἀπὸ τῶν ΖΓ, ΓΚ angle. So, for the same (reasons), the angles at B and
τοῖς ἀπὸ τῶν ΖΒ, ΒΚ ἐστιν ἴσα, ὧν τὸ ἀπὸ τῆς ΖΓ τῷ ἀπὸ D are also right-angles. And since angle FCK is a right-
τῆς ΖΒ ἐστιν ἴσον· λοιπὸν ἄρα τὸ ἀπὸ τῆς ΓΚ τῷ ἀπὸ τῆς angle, the (square) on FK is thus equal to the (sum of
ΒΚ ἐστιν ἴσον. ἴση ἄρα ἡ ΒΚ τῇ ΓΚ. καὶ ἐπεὶ ἴση ἐστὶν the squares) on FC and CK [Prop. 1.47]. So, for the
ἡ ΖΒ τῇ ΖΓ, καὶ κοινὴ ἡ ΖΚ, δύο δὴ αἱ ΒΖ, ΖΚ δυσὶ ταῖς same (reasons), the (square) on FK is also equal to the
ΓΖ, ΖΚ ἴσαι εἰσίν· καὶ βάσις ἡ ΒΚ βάσει τῇ ΓΚ [ἐστιν] ἴση· (sum of the squares) on FB and BK. So that the (sum
γωνία ἄρα ἡ μὲν ὑπὸ ΒΖΚ [γωνίᾳ] τῇ ὑπὸ ΚΖΓ ἐστιν ἴση· of the squares) on FC and CK is equal to the (sum of
ἡ δὲ ὑπὸ ΒΚΖ τῇ ὑπὸ ΖΚΓ· διπλῆ ἄρα ἡ μὲν ὑπὸ ΒΖΓ τῆς the squares) on FB and BK, of which the (square) on
ὑπὸ ΚΖΓ, ἡ δὲ ὑπὸ ΒΚΓ τῆς ὑπὸ ΖΚΓ. διὰ τὰ αὐτὰ δὴ καὶ FC is equal to the (square) on FB. Thus, the remain-
ἡ μὲν ὑπὸ ΓΖΔ τῆς ὑπὸ ΓΖΛ ἐστι διπλῆ, ἡ δὲ ὑπὸ ΔΛΓ ing (square) on CK is equal to the remaining (square)
τῆς ὑπὸ ΖΛΓ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΓ περιφέρεια τῇ ΓΔ, on BK. Thus, BK (is) equal to CK. And since FB is
ἴση ἐστὶ καὶ γωνία ἡ ὑπὸ ΒΖΓ τῇ ὑπὸ ΓΖΔ. καί ἐστιν ἡ equal to FC, and FK (is) common, the two (straight-
μὲν ὑπὸ ΒΖΓ τῆς ὑπὸ ΚΖΓ διπλῆ, ἡ δὲ ὑπὸ ΔΖΓ τῆς ὑπὸ lines) BF, FK are equal to the two (straight-lines) CF,
ΛΖΓ· ἴση ἄρα καὶ ἡ ὑπὸ ΚΖΓ τῇ ὑπὸ ΛΖΓ· ἐστὶ δὲ καὶ ἡ FK. And the base BK [is] equal to the base CK. Thus,
ὑπὸ ΖΓΚ γωνία τῇ ὑπὸ ΖΓΛ ἴση. δύο δὴ τρίγωνά ἐστι τὰ angle BFK is equal to [angle] KFC [Prop. 1.8]. And
ΖΚΓ, ΖΛΓ τὰς δύο γωνίας ταῖς δυσὶ γωνίαις ἴσας ἔχοντα BKF (is equal) to FKC [Prop. 1.8]. Thus, BFC (is)
καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην κοινὴν αὐτῶν τὴν ΖΓ· double KFC, and BKC (is double) FKC. So, for the
καὶ τὰς λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξει same (reasons), CFD is also double CFL, and DLC (is
καὶ τὴν λοιπὴν γωνίαν τῇ λοιπῇ γωνίᾳ· ἴση ἄρα ἡ μὲν ΚΓ also double) FLC. And since circumference BC is equal
εὐθεῖα τῇ ΓΛ, ἡ δὲ ὑπὸ ΖΚΓ γωνία τῇ ὑπὸ ΖΛΓ. καὶ ἐπεὶ ἴση to CD, angle BFC is also equal to CFD [Prop. 3.27].
ἐστὶν ἡ ΚΓ τῇ ΓΛ, διπλῆ ἄρα ἡ ΚΛ τῆς ΚΓ. διὰ τὰ αὐτα δὴ And BFC is double KFC, and DFC (is double) LFC.
δειχθήσεται καὶ ἡ ΘΚ τῆς ΒΚ διπλῆ. καί ἐστιν ἡ ΒΚ τῇ ΚΓ Thus, KFC is also equal to LFC. And angle FCK is also
ἴση· καὶ ἡ ΘΚ ἄρα τῇ ΚΛ ἐστιν ἴση. ὁμοίως δὴ δειχθήσεται equal to FCL. So, FKC and FLC are two triangles hav-

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καὶ ἑκάστη τῶν ΘΗ, ΗΜ, ΜΛ ἑκατέρᾳ τῶν ΘΚ, ΚΛ ἴση· ing two angles equal to two angles, and one side equal
ἰσόπλευρον ἄρα ἐστὶ τὸ ΗΘΚΛΜ πεντάγωνον. λέγω δή, to one side, (namely) their common (side) FC. Thus,
ὅτι καὶ ἰσογώνιον. ἐπεὶ γὰρ ἴση ἐστὶν ἡ ὑπὸ ΖΚΓ γωνία τῇ they will also have the remaining sides equal to the (cor-
ὑπὸ ΖΛΓ, καὶ ἐδείχθη τῆς μὲν ὑπὸ ΖΚΓ διπλῆ ἡ ὑπὸ ΘΚΛ, responding) remaining sides, and the remaining angle to
τῆς δὲ ὑπὸ ΖΛΓ διπλῆ ἡ ὑπὸ ΚΛΜ, καὶ ἡ ὑπὸ ΘΚΛ ἄρα the remaining angle [Prop. 1.26]. Thus, the straight-line
τῇ ὑπὸ ΚΛΜ ἐστιν ἴση. ὁμοίως δὴ δειχθήσεται καὶ ἑκάστη KC (is) equal to CL, and the angle FKC to FLC. And
τῶν ὑπὸ ΚΘΗ, ΘΗΜ, ΗΜΛ ἑκατέρᾳ τῶν ὑπὸ ΘΚΛ, ΚΛΜ since KC is equal to CL, KL (is) thus double KC. So,
ἴση· αἱ πέντε ἄρα γωνίαι αἱ ὑπὸ ΗΘΚ, ΘΚΛ, ΚΛΜ, ΛΜΗ, for the same (reasons), it can be shown that HK (is) also
ΜΗΘ ἴσαι ἀλλήλαις εἰσίν. ἰσογώνιον ἄρα ἐστὶ τὸ ΗΘΚΛΜ double BK. And BK is equal to KC. Thus, HK is also
πεντάγωνον. ἐδείχθη δὲ καὶ ἰσόπλευρον, καὶ περιγέγραπται equal to KL. So, similarly, each of HG, GM, and ML
περὶ τὸν ΑΒΓΔΕ κύκλον. can also be shown (to be) equal to each of HK and KL.
[Περὶ τὸν δοθέντα ἄρα κύκλον πεντάγωνον ἰσόπλευρόν Thus, pentagon GHKLM is equilateral. So I say that
τε καὶ ἰσογώνιον περιγέγραπται]· ὅπερ ἔδει ποιῆσαι. (it is) also equiangular. For since angle FKC is equal to
FLC, and HKL was shown (to be) double FKC, and
KLM double FLC, HKL is thus also equal to KLM.
So, similarly, each of KHG, HGM, and GML can also
be shown (to be) equal to each of HKL and KLM. Thus,
the five angles GHK, HKL, KLM, LMG, and MGH
igþ and has been circumscribed about circle ABCDE.
are equal to one another. Thus, the pentagon GHKLM
is equiangular. And it was also shown (to be) equilateral,

[Thus, an equilateral and equiangular pentagon has
been circumscribed about the given circle]. (Which is)
the very thing it was required to do.

† See the footnote to Prop. 3.34.

.
Proposition 13
Εἰς τὸ δοθὲν πεντάγωνον, ὅ ἐστιν ἰσόπλευρόν τε καὶ To inscribe a circle in a given pentagon, which is equi-
ἰσογώνιον, κύκλον ἐγγράψαι. lateral and equiangular.
Α A

Η Μ G M


Β Ε B E
Ζ F



Θ Λ H L



Γ Κ ∆ C K D
῎Εστω τὸ δοθὲν πεντάγωνον ἰσόπλευρόν τε καὶ ἰσογώνι- Let ABCDE be the given equilateral and equiangular
ον τὸ ΑΒΓΔΕ· δεῖ δὴ εἰς τὸ ΑΒΓΔΕ πεντάγωνον κύκλον pentagon. So it is required to inscribe a circle in pentagon
ἐγγράψαι. ABCDE.
Τετμήσθω γὰρ ἑκατέρα τῶν ὑπὸ ΒΓΔ, ΓΔΕ γωνιῶν For let angles BCD and CDE have each been cut
δίχα ὑπὸ ἑκατέρας τῶν ΓΖ, ΔΖ εὐθειῶν· καὶ ἀπὸ τοῦ Ζ in half by each of the straight-lines CF and DF (re-
σημείου, καθ᾿ ὃ συμβάλλουσιν ἀλλήλαις αἱ ΓΖ, ΔΖ εὐθεῖαι, spectively) [Prop. 1.9]. And from the point F, at which
ἐπεζεύχθωσαν αἱ ΖΒ, ΖΑ, ΖΕ εὐθεῖαι. καὶ ἐπεὶ ἴση ἐστὶν the straight-lines CF and DF meet one another, let the


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ἡ ΒΓ τῇ ΓΔ, κοινὴ δὲ ἡ ΓΖ, δύο δὴ αἱ ΒΓ, ΓΖ δυσὶ ταῖς straight-lines FB, FA, and FE have been joined. And
ΔΓ, ΓΖ ἴσαι εἰσίν· καὶ γωνία ἡ ὑπὸ ΒΓΖ γωνίᾳ τῇ ὑπὸ since BC is equal to CD, and CF (is) common, the two
ΔΓΖ [ἐστιν] ἴση· βάσις ἄρα ἡ ΒΖ βάσει τῇ ΔΖ ἐστιν ἴση, (straight-lines) BC, CF are equal to the two (straight-
καὶ τὸ ΒΓΖ τρίγωνον τῷ ΔΓΖ τριγώνῳ ἐστιν ἴσον, καὶ αἱ lines) DC, CF. And angle BCF [is] equal to angle
λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται, ὑφ᾿ ἃς αἱ DCF. Thus, the base BF is equal to the base DF, and
ἴσαι πλευραὶ ὑποτείνουσιν· ἴση ἄρα ἡ ὑπὸ ΓΒΖ γωνία τῇ triangle BCF is equal to triangle DCF, and the remain-
ὑπὸ ΓΔΖ. καὶ ἐπεὶ διπλῆ ἐστιν ἡ ὑπὸ ΓΔΕ τῆς ὑπὸ ΓΔΖ, ing angles will be equal to the (corresponding) remain-
ἴση δὲ ἡ μὲν ὑπὸ ΓΔΕ τῇ ὑπὸ ΑΒΓ, ἡ δὲ ὑπὸ ΓΔΖ τῇ ὑπὸ ing angles which the equal sides subtend [Prop. 1.4].
ΓΒΖ, καὶ ἡ ὑπὸ ΓΒΑ ἄρα τῆς ὑπὸ ΓΒΖ ἐστι διπλῆ· ἴση Thus, angle CBF (is) equal to CDF. And since CDE
ἄρα ἡ ὑπὸ ΑΒΖ γωνία τῇ ὑπὸ ΖΒΓ· ἡ ἄρα ὑπὸ ΑΒΓ γωνία is double CDF, and CDE (is) equal to ABC, and CDF
δίχα τέτμηται ὑπὸ τῆς ΒΖ εὐθείας. ὁμοίως δὴ δειχθήσεται, to CBF, CBA is thus also double CBF. Thus, angle
ὅτι καὶ ἑκατέρα τῶν ὑπὸ ΒΑΕ, ΑΕΔ δίχα τέτμηται ὑπὸ ABF is equal to FBC. Thus, angle ABC has been cut
ἑκατέρας τῶν ΖΑ, ΖΕ εὐθειῶν. ἤχθωσαν δὴ ἀπὸ τοῦ Ζ in half by the straight-line BF. So, similarly, it can be
σημείου ἐπὶ τὰς ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΑ εὐθείας κάθετοι shown that BAE and AED have been cut in half by the
αἱ ΖΗ, ΖΘ, ΖΚ, ΖΛ, ΖΜ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΘΓΖ straight-lines FA and FE, respectively. So let FG, FH,
γωνία τῇ ὑπὸ ΚΓΖ, ἐστὶ δὲ καὶ ὀρθὴ ἡ ὑπὸ ΖΘΓ [ὀρθῇ] FK, FL, and FM have been drawn from point F, per-
τῇ ὑπὸ ΖΚΓ ἴση, δύο δὴ τρίγωνά ἐστι τὰ ΖΘΓ, ΖΚΓ τὰς pendicular to the straight-lines AB, BC, CD, DE, and
δύο γωνίας δυσὶ γωνίαις ἴσας ἔχοντα καὶ μίαν πλευρὰν μιᾷ EA (respectively) [Prop. 1.12]. And since angle HCF
πλευρᾷ ἴσην κοινὴν αὐτῶν τὴν ΖΓ ὑποτείνουσαν ὑπὸ μίαν is equal to KCF, and the right-angle FHC is also equal
τῶν ἴσων γωνιῶν· καὶ τὰς λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς to the [right-angle] FKC, FHC and FKC are two tri-
πλευραῖς ἴσας ἕξει· ἴση ἄρα ἡ ΖΘ κάθετος τῂ ΖΚ καθέτῳ. angles having two angles equal to two angles, and one
ὁμοίως δὴ δειχθήσεται, ὅτι καὶ ἑκάστη τῶν ΖΛ, ΖΜ, ΖΗ side equal to one side, (namely) their common (side) FC,
ἑκατέρᾳ τῶν ΖΘ, ΖΚ ἴση ἐστίν· αἱ πέντε ἄρα εὐθεῖαι αἱ ΖΗ, subtending one of the equal angles. Thus, they will also
ΖΘ, ΖΚ, ΖΛ, ΖΜ ἴσαι ἀλλήλαις εἰσίν. ὁ ἄρα κέντρῳ τῷ Ζ have the remaining sides equal to the (corresponding)
διαστήματι δὲ ἑνὶ τῶν Η, Θ, Κ, Λ, Μ κύκλος γραφόμενος remaining sides [Prop. 1.26]. Thus, the perpendicular
ἥξει καὶ διὰ τῶν λοιπῶν σημείων καὶ ἐφάψεται τῶν ΑΒ, ΒΓ, FH (is) equal to the perpendicular FK. So, similarly, it
ΓΔ, ΔΕ, ΕΑ εὐθειῶν διὰ τὸ ὀρθὰς εἶναι τὰς πρὸς τοῖς Η, can be shown that FL, FM, and FG are each equal to
Θ, Κ, Λ, Μ σημείοις γωνίας. εἰ γὰρ οὐκ ἐφάψεται αὐτῶν, each of FH and FK. Thus, the five straight-lines FG,
ἀλλὰ τεμεῖ αὐτάς, συμβήσεται τὴν τῇ διαμέτρῳ τοῦ κύκλου FH, FK, FL, and FM are equal to one another. Thus,
πρὸς ὀρθὰς ἀπ᾿ ἄκρας ἀγομένην ἐντὸς πίπτειν τοῦ κύκλου· the circle drawn with center F, and radius one of G, H,
ὅπερ ἄτοπον ἐδείχθη. οὐκ ἄρα ὁ κέντρῳ τῷ Ζ διαστήματι δὲ K, L, or M, will also go through the remaining points,
ἑνὶ τῶν Η, Θ, Κ, Λ, Μ σημείων γραφόμενος κύκλος τεμεῖ and will touch the straight-lines AB, BC, CD, DE, and
τὰς ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΑ εὐθείας· ἐφάψεται ἄρα αὐτῶν. EA, on account of the angles at points G, H, K, L, and
γεγράφθω ὡς ὁ ΗΘΚΛΜ. M being right-angles. For if it does not touch them, but
Εἰς ἄρα τὸ δοθὲν πεντάγωνον, ὅ ἐστιν ἰσόπλευρόν τε cuts them, it follows that a (straight-line) drawn at right-
καὶ ἰσογώνιον, κύκλος ἐγγέγραπται· ὅπερ ἔδει ποιῆσαι. angles to the diameter of the circle, from its extremity,
falls inside the circle. The very thing was shown (to be)
idþ the straight-lines AB, BC, CD, DE, or EA. Thus, it will
absurd [Prop. 3.16]. Thus, the circle drawn with center
F, and radius one of G, H, K, L, or M, does not cut

touch them. Let it have been drawn, like GHKLM (in
the figure).
Thus, a circle has been inscribed in the given pen-
tagon which is equilateral and equiangular. (Which is)
the very thing it was required to do.

.
Proposition 14
Περὶ τὸ δοθὲν πεντάγωνον, ὅ ἐστιν ἰσόπλευρόν τε καὶ To circumscribe a circle about a given pentagon which
ἰσογώνιον, κύκλον περιγράψαι. is equilateral and equiangular.
῎Εστω τὸ δοθὲν πεντάγωνον, ὅ ἐστιν ἰσόπλευρόν τε καὶ Let ABCDE be the given pentagon which is equilat-


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ELEMENTS BOOK 4



ἰσογώνιον, τὸ ΑΒΓΔΕ· δεῖ δὴ περὶ τὸ ΑΒΓΔΕ πεντάγωνον eral and equiangular. So it is required to circumscribe a
κύκλον περιγράψαι. circle about the pentagon ABCDE.
Α A





Β Ε B E
Ζ F







Γ ∆ C D

Τετμήσθω δὴ ἑκατέρα τῶν ὑπὸ ΒΓΔ, ΓΔΕ γωνιῶν So let angles BCD and CDE have been cut in half by
δίχα ὑπὸ ἑκατέρας τῶν ΓΖ, ΔΖ, καὶ ἀπὸ τοῦ Ζ σημείου, the (straight-lines) CF and DF, respectively [Prop. 1.9].
καθ᾿ ὃ συμβάλλουσιν αἱ εὐθεῖαι, ἐπὶ τὰ Β, Α, Ε σημεῖα And let the straight-lines FB, FA, and FE have been
ἐπεζεύχθωσαν εὐθεῖαι αἱ ΖΒ, ΖΑ, ΖΕ. ὁμοίως δὴ τῷ πρὸ joined from point F, at which the straight-lines meet,
τούτου δειχθήσεται, ὅτι καὶ ἑκάστη τῶν ὑπὸ ΓΒΑ, ΒΑΕ, to the points B, A, and E (respectively). So, similarly,
ΑΕΔ γωνιῶν δίχα τέτμηται ὑπὸ ἑκάστης τῶν ΖΒ, ΖΑ, ΖΕ to the (proposition) before this (one), it can be shown
εὐθειῶν. καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΒΓΔ γωνία τῇ ὑπὸ ΓΔΕ, that angles CBA, BAE, and AED have also been cut
καί ἐστι τῆς μὲν ὑπὸ ΒΓΔ ἡμίσεια ἡ ὑπὸ ΖΓΔ, τῆς δὲ ὑπὸ in half by the straight-lines FB, FA, and FE, respec-
ΓΔΕ ἡμίσεια ἡ ὑπὸ ΓΔΖ, καὶ ἡ ὑπὸ ΖΓΔ ἄρα τῇ ὑπὸ ΖΔΓ tively. And since angle BCD is equal to CDE, and FCD
ἐστιν ἴση· ὥστε καὶ πλευρὰ ἡ ΖΓ πλευρᾷ τῇ ΖΔ ἐστιν ἴση. is half of BCD, and CDF half of CDE, FCD is thus
ὁμοίως δὴ δειχθήσεται, ὅτι καὶ ἑκάστη τῶν ΖΒ, ΖΑ, ΖΕ also equal to FDC. So that side FC is also equal to side
ἑκατέρᾳ τῶν ΖΓ, ΖΔ ἐστιν ἴση· αἱ πέντε ἄρα εὐθεῖαι αἱ FD [Prop. 1.6]. So, similarly, it can be shown that FB,
ΖΑ, ΖΒ, ΖΓ, ΖΔ, ΖΕ ἴσαι ἀλλήλαις εἰσίν. ὀ ἄρα κέντρῳ FA, and FE are also each equal to each of FC and FD.
τῷ Ζ καὶ διαστήματι ἑνὶ τῶν ΖΑ, ΖΒ, ΖΓ, ΖΔ, ΖΕ κύκλος Thus, the five straight-lines FA, FB, FC, FD, and FE
ieþ and let it be ABCDE.
γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν σημείων καὶ ἔσται πε- are equal to one another. Thus, the circle drawn with
ριγεγραμμένος. περιγεγράφθω καὶ ἔστω ὁ ΑΒΓΔΕ. center F, and radius one of FA, FB, FC, FD, or FE,
Περὶ ἄρα τὸ δοθὲν πεντάγωνον, ὅ ἐστιν ἰσόπλευρόν τε will also go through the remaining points, and will have
καὶ ἰσογώνιον, κύκλος περιγέγραπται· ὅπερ ἔδει ποιῆσαι. been circumscribed. Let it have been (so) circumscribed,

Thus, a circle has been circumscribed about the given
pentagon, which is equilateral and equiangular. (Which
is) the very thing it was required to do.

Proposition 15
.
Εἰς τὸν δοθέντα κύκλον ἑξάγωνον ἰσόπλευρόν τε καὶ To inscribe an equilateral and equiangular hexagon in
ἰσογώνιον ἐγγράψαι. a given circle.
῎Εστω ὁ δοθεὶς κύκλος ὁ ΑΒΓΔΕΖ· δεῖ δὴ εἰς τὸν Let ABCDEF be the given circle. So it is required to
ΑΒΓΔΕΖ κύκλον ἑξάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον inscribe an equilateral and equiangular hexagon in circle
ἐγγράψαι. ABCDEF.
῎Ηχθω τοῦ ΑΒΓΔΕΖ κύκλου διάμετρος ἡ ΑΔ, καὶ Let the diameter AD of circle ABCDEF have been
†
εἰλήφθω τὸ κέντρον τοῦ κύκλου τὸ Η, καὶ κέντρῳ μὲν drawn, and let the center G of the circle have been
τῷ Δ διαστήματι δὲ τῷ ΔΗ κύκλος γεγράφθω ὁ ΕΗΓΘ, found [Prop. 3.1]. And let the circle EGCH have been
καὶ ἐπιζευχθεῖσαι αἱ ΕΗ, ΓΗ διήχθωσαν ἐπὶ τὰ Β, Ζ σημεῖα, drawn, with center D, and radius DG. And EG and CG
καὶ ἐπεζεύχθωσαν αἱ ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΖ, ΖΑ· λέγω, ὅτι being joined, let them have been drawn across (the cir-



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τὸ ΑΒΓΔΕΖ ἑξάγωνον ἰσόπλευρόν τέ ἐστι καὶ ἰσογώνιον. cle) to points B and F (respectively). And let AB, BC,
CD, DE, EF, and FA have been joined. I say that the
hexagon ABCDEF is equilateral and equiangular.
Θ H




∆ D


Γ Ε C E
Η G



Β Ζ B F


Α A
᾿Επεὶ γὰρ τὸ Η σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓΔΕΖ For since point G is the center of circle ABCDEF,
κύκλου, ἴση ἐστὶν ἡ ΗΕ τῇ ΗΔ. πάλιν, ἐπεὶ τὸ Δ σημεῖον GE is equal to GD. Again, since point D is the cen-
κέντρον ἐστὶ τοῦ ΗΓΘ κύκλου, ἴση ἐστὶν ἡ ΔΕ τῇ ΔΗ. ter of circle GCH, DE is equal to DG. But, GE was
ἀλλ᾿ ἡ ΗΕ τῇ ΗΔ ἐδείχθη ἴση· καὶ ἡ ΗΕ ἄρα τῇ ΕΔ ἴση shown (to be) equal to GD. Thus, GE is also equal to
ἐστίν· ἰσόπλευρον ἄρα ἐστὶ τὸ ΕΗΔ τρίγωνον· καὶ αἱ τρεῖς ED. Thus, triangle EGD is equilateral. Thus, its three
ἄρα αὐτοῦ γωνίαι αἱ ὑπὸ ΕΗΔ, ΗΔΕ, ΔΕΗ ἴσαι ἀλλήλαις angles EGD, GDE, and DEG are also equal to one an-
εἰσίν, ἐπειδήπερ τῶν ἰσοσκελῶν τριγώνων αἱ πρὸς τῇ βάσει other, inasmuch as the angles at the base of isosceles tri-
γωνίαι ἴσαι ἀλλήλαις εἰσίν· καί εἰσιν αἱ τρεῖς τοῦ τριγώνου angles are equal to one another [Prop. 1.5]. And the
γωνίαι δυσὶν ὀρθαῖς ἴσαι· ἡ ἄρα ὑπὸ ΕΗΔ γωνία τρίτον ἐστὶ three angles of the triangle are equal to two right-angles
δύο ὀρθῶν. ὁμοίως δὴ δειχθήσεται καὶ ἡ ὑπὸ ΔΗΓ τρίτον [Prop. 1.32]. Thus, angle EGD is one third of two right-
δύο ὀρθῶν. καὶ ἐπεὶ ἡ ΓΗ εὐθεῖα ἐπὶ τὴν ΕΒ σταθεῖσα τὰς angles. So, similarly, DGC can also be shown (to be)
ἐφεξῆς γωνίας τὰς ὑπὸ ΕΗΓ, ΓΗΒ δυσὶν ὀρθαῖς ἴσας ποιεῖ, one third of two right-angles. And since the straight-line
καὶ λοιπὴ ἄρα ἡ ὑπὸ ΓΗΒ τρίτον ἐστὶ δύο ὀρθῶν· αἱ ἄρα CG, standing on EB, makes adjacent angles EGC and
ὑπὸ ΕΗΔ, ΔΗΓ, ΓΗΒ γωνίαι ἴσαι ἀλλήλαις εἰσίν· ὥστε καὶ CGB equal to two right-angles [Prop. 1.13], the remain-
αἱ κατὰ κορυφὴν αὐταῖς αἱ ὑπὸ ΒΗΑ, ΑΗΖ, ΖΗΕ ἴσαι εἰσὶν ing angle CGB is thus also one third of two right-angles.
[ταῖς ὑπὸ ΕΗΔ, ΔΗΓ, ΓΗΒ]. αἱ ἓξ ἄρα γωνίαι αἱ ὑπὸ ΕΗΔ, Thus, angles EGD, DGC, and CGB are equal to one an-
ΔΗΓ, ΓΗΒ, ΒΗΑ, ΑΗΖ, ΖΗΕ ἴσαι ἀλλήλαις εἰσίν. αἱ δὲ other. And hence the (angles) opposite to them BGA,
ἴσαι γωνίαι ἐπὶ ἴσων περιφερειῶν βεβήκασιν· αἱ ἓξ ἄρα πε- AGF, and FGE are also equal [to EGD, DGC, and
ριφέρειαι αἱ ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΖ, ΖΑ ἴσαι ἀλλήλαις εἰσίν. CGB (respectively)] [Prop. 1.15]. Thus, the six angles
ὑπὸ δὲ τὰς ἴσας περιφερείας αἱ ἴσαι εὐθεῖαι ὑποτείνουσιν· EGD, DGC, CGB, BGA, AGF, and FGE are equal
αἱ ἓξ ἄρα εὐθεῖαι ἴσαι ἀλλήλαις εἰσίν· ἰσόπλευρον ἄρα ἐστὶ to one another. And equal angles stand on equal cir-
το ΑΒΓΔΕΖ ἑξάγωνον. λέγω δή, ὅτι καὶ ἰσογώνιον. ἐπεὶ cumferences [Prop. 3.26]. Thus, the six circumferences
γὰρ ἴση ἐστὶν ἡ ΖΑ περιφέρεια τῇ ΕΔ περιφερείᾳ, κοινὴ AB, BC, CD, DE, EF, and FA are equal to one an-
προσκείσθω ἡ ΑΒΓΔ περιφέρεια· ὅλη ἄρα ἡ ΖΑΒΓΔ ὅλῃ other. And equal circumferences are subtended by equal
τῇ ΕΔΓΒΑ ἐστιν ἴση· καὶ βέβηκεν ἐπὶ μὲν τῆς ΖΑΒΓΔ straight-lines [Prop. 3.29]. Thus, the six straight-lines
περιφερείας ἡ ὑπὸ ΖΕΔ γωνία, ἐπὶ δὲ τῆς ΕΔΓΒΑ περι- (AB, BC, CD, DE, EF, and FA) are equal to one
φερείας ἡ ὑπὸ ΑΖΕ γωνία· ἴση ἄρα ἡ ὑπὸ ΑΖΕ γωνία τῇ another. Thus, hexagon ABCDEF is equilateral. So,
ὑπὸ ΔΕΖ. ὁμοίως δὴ δειχθήσεται, ὅτι καὶ αἱ λοιπαὶ γωνίαι I say that (it is) also equiangular. For since circumfer-
τοῦ ΑΒΓΔΕΖ ἑξαγώνου κατὰ μίαν ἴσαι εἰσὶν ἑκατέρᾳ τῶν ence FA is equal to circumference ED, let circumference
ὑπὸ ΑΖΕ, ΖΕΔ γωνιῶν· ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓΔΕΖ ABCD have been added to both. Thus, the whole of
ἑξάγωνον. ἐδείχθη δὲ καὶ ἰσόπλευρον· καὶ ἐγγέγραπται εἰς FABCD is equal to the whole of EDCBA. And angle
τὸν ΑΒΓΔΕΖ κύκλον. FED stands on circumference FABCD, and angle AFE
Εἰς ἄρα τὸν δοθέντα κύκλον ἑξάγωνον ἰσόπλευρόν τε on circumference EDCBA. Thus, angle AFE is equal


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καὶ ἰσογώνιον ἐγγέγραπται· ὅπερ ἔδει ποιῆσαι. to DEF [Prop. 3.27]. Similarly, it can also be shown
ìri sma Thus, hexagon ABCDEF is equiangular. And it was also
that the remaining angles of hexagon ABCDEF are in-
dividually equal to each of the angles AFE and FED.

shown (to be) equilateral. And it has been inscribed in
circle ABCDE.
Thus, an equilateral and equiangular hexagon has
been inscribed in the given circle. (Which is) the very
thing it was required to do.
.
Corollary
᾿Εκ δὴ τούτου φανερόν, ὅτι ἡ τοῦ ἑξαγώνου πλευρὰ ἴση So, from this, (it is) manifest that a side of the
ἐστὶ τῇ ἐκ τοῦ κέντρου τοῦ κύκλου. hexagon is equal to the radius of the circle.
῾Ομοίως δὲ τοῖς ἐπὶ τοῦ πενταγώνου ἐὰν διὰ τῶν κατὰ And similarly to a pentagon, if we draw tangents
τὸν κύκλον διαιρέσεων ἐφαπτομένας τοῦ κύκλου ἀγάγωμεν, to the circle through the (sixfold) divisions of the (cir-
iþ hexagon. (Which is) the very thing it was required to do.
περιγραφήσεται περὶ τὸν κύκλον ἑξάγωνον ἰσόπλευρόν cumference of the) circle, an equilateral and equiangu-
τε καὶ ἰσογώνιον ἀκολούθως τοῖς ἐπὶ τοῦ πενταγώνου lar hexagon can be circumscribed about the circle, analo-
εἰρημένοις. καὶ ἔτι διὰ τῶν ὁμοίων τοῖς ἐπὶ τοῦ πενταγώνου gously to the aforementioned pentagon. And, further, by
εἰρημένοις εἰς τὸ δοθὲν ἑξάγωνον κύκλον ἐγγράψομέν τε (means) similar to the aforementioned pentagon, we can
καὶ περιγράψομεν· ὅπερ ἔδει ποιῆσαι. inscribe and circumscribe a circle in (and about) a given


† See the footnote to Prop. 4.6.

Proposition 16
.
Εἰς τὸν δοθέντα κύκλον πεντεκαιδεκάγωνον ἰσόπλευρόν To inscribe an equilateral and equiangular fifteen-
τε καὶ ἰσογώνιον ἐγγράψαι. sided figure in a given circle.
Α A




Β B



Ε E

Γ ∆ C D




῎Εστω ὁ δοθεὶς κύκλος ὁ ΑΒΓΔ· δεῖ δὴ εἰς τὸν ΑΒΓΔ Let ABCD be the given circle. So it is required to in-
κύκλον πεντεκαιδεκάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον scribe an equilateral and equiangular fifteen-sided figure
ἐγγράψαι. in circle ABCD.
᾿Εγγεγράφθω εἰς τὸν ΑΒΓΔ κύκλον τριγώνου μὲν ἰσο- Let the side AC of an equilateral triangle inscribed
πλεύρου τοῦ εἰς αὐτὸν ἐγγραφομένου πλευρὰ ἡ ΑΓ, πεν- in (the circle) [Prop. 4.2], and (the side) AB of an (in-
ταγώνου δὲ ἰσοπλεύρου ἡ ΑΒ· οἵων ἄρα ἐστὶν ὁ ΑΒΓΔ scribed) equilateral pentagon [Prop. 4.11], have been in-
κύκλος ἴσων τμήματων δεκαπέντε, τοιούτων ἡ μὲν ΑΒΓ scribed in circle ABCD. Thus, just as the circle ABCD
περιφέρεια τρίτον οὖσα τοῦ κύκλου ἔσται πέντε, ἡ δὲ ΑΒ is (made up) of fifteen equal pieces, the circumference
περιφέρεια πέμτον οὖσα τοῦ κύκλου ἔσται τριῶν· λοιπὴ ἄρα ABC, being a third of the circle, will be (made up) of five


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ἡ ΒΓ τῶν ἴσων δύο. τετμήσθω ἡ ΒΓ δίχα κατὰ τὸ Ε· such (pieces), and the circumference AB, being a fifth of
ἑκατέρα ἄρα τῶν ΒΕ, ΕΓ περιφερειῶν πεντεκαιδέκατόν ἐστι the circle, will be (made up) of three. Thus, the remain-
τοῦ ΑΒΓΔ κύκλου. der BC (will be made up) of two equal (pieces). Let (cir-
᾿Εὰν ἄρα ἐπιζεύξαντες τὰς ΒΕ, ΕΓ ἴσας αὐταῖς κατὰ τὸ cumference) BC have been cut in half at E [Prop. 3.30].
συνεχὲς εὐθείας ἐναρμόσωμεν εἰς τὸν ΑΒΓΔ[Ε] κύκλον, Thus, each of the circumferences BE and EC is one fif-
ἔσται εἰς αὐτὸν ἐγγεγραμμένον πεντεκαιδεκάγωνον ἰσόπλευ- teenth of the circle ABCDE.
ρόν τε καὶ ἰσογώνιον· ὅπερ ἔδει ποιῆσαι. Thus, if, joining BE and EC, we continuously in-
῾Ομοίως δὲ τοῖς ἐπὶ τοῦ πενταγώνου ἐὰν διὰ τῶν sert straight-lines equal to them into circle ABCD[E]
κατὰ τὸν κύκλον διαιρέσεων ἐφαπτομένας τοῦ κύκλου [Prop. 4.1], then an equilateral and equiangular fifteen-
ἀγάγωμεν, περιγραφήσεται περὶ τὸν κύκλον πεντεκαι- sided figure will have been inserted into (the circle).
δεκάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον. ἔτι δὲ διὰ (Which is) the very thing it was required to do.
τῶν ὁμοίων τοῖς ἐπὶ τοῦ πενταγώνου δείξεων καὶ εἰς τὸ And similarly to the pentagon, if we draw tangents to
δοθὲν πεντεκαιδεκάγωνον κύκλον ἐγγράψομέν τε καὶ πε- the circle through the (fifteenfold) divisions of the (cir-
ριγράψομεν· ὅπερ ἔδει ποιῆσαι. cumference of the) circle, we can circumscribe an equilat-
eral and equiangular fifteen-sided figure about the circle.
And, further, through similar proofs to the pentagon, we
can also inscribe and circumscribe a circle in (and about)
a given fifteen-sided figure. (Which is) the very thing it
was required to do.


















































127

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Proportion †
































































† The theory of proportion set out in this book is generally attributed to Eudoxus of Cnidus. The novel feature of this theory is its ability to deal
with irrational magnitudes, which had hitherto been a major stumbling block for Greek mathematicians. Throughout the footnotes in this book,
α, β, γ, etc., denote general (possibly irrational) magnitudes, whereas m, n, l, etc., denote positive integers.

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.
Definitions
αʹ. Μέρος ἐστὶ μέγεθος μεγέθους τὸ ἔλασσον τοῦ 1. A magnitude is a part of a(nother) magnitude, the
μείζονος, ὅταν καταμετρῇ τὸ μεῖζον. lesser of the greater, when it measures the greater. †
βʹ. Πολλαπλάσιον δὲ τὸ μεῖζον τοῦ ἐλάττονος, ὅταν κα- 2. And the greater (magnitude is) a multiple of the
ταμετρῆται ὑπὸ τοῦ ἐλάττονος. lesser when it is measured by the lesser.
γʹ. Λόγος ἐστὶ δύο μεγεθῶν ὁμογενῶν ἡ κατὰ πη- 3. A ratio is a certain type of condition with respect to
λικότητά ποια σχέσις. size of two magnitudes of the same kind. ‡
δʹ. Λόγον ἔχειν πρὸς ἄλληλα μεγέθη λέγεται, ἃ δύναται 4. (Those) magnitudes are said to have a ratio with re-
πολλαπλασιαζόμενα ἀλλήλων ὑπερέχειν. spect to one another which, being multiplied, are capable
εʹ. ᾿Εν τῷ αὐτῷ λόγῳ μεγέθη λέγεται εἶναι πρῶτον of exceeding one another. §
πρὸς δεύτερον καὶ τρίτον πρὸς τέταρτον, ὅταν τὰ τοῦ 5. Magnitudes are said to be in the same ratio, the first
πρώτου καί τρίτου ἰσάκις πολλαπλάσια τῶν τοῦ δευτέρου to the second, and the third to the fourth, when equal
καὶ τετάρτου ἰσάκις πολλαπλασίων καθ᾿ ὁποιονοῦν πολλα- multiples of the first and the third either both exceed, are
πλασιασμὸν ἑκάτερον ἑκατέρου ἢ ἅμα ὑπερέχῃ ἢ ἅμα ἴσα ᾖ both equal to, or are both less than, equal multiples of the
ἢ ἅμα ἐλλείπῇ ληφθέντα κατάλληλα. second and the fourth, respectively, being taken in corre-
ϛʹ. Τὰ δὲ τὸν αὐτὸν ἔχοντα λόγον μεγέθη ἀνάλογον sponding order, according to any kind of multiplication
καλείσθω. whatever. ¶
ζʹ. ῞Οταν δὲ τῶν ἰσάκις πολλαπλασίων τὸ μὲν τοῦ 6. And let magnitudes having the same ratio be called
πρώτου πολλαπλάσιον ὑπερέχῃ τοῦ τοῦ δευτέρου πολ- proportional. ∗
λαπλασίου, τὸ δὲ τοῦ τρίτου πολλαπλάσιον μὴ ὑπερέχῃ 7. And when for equal multiples (as in Def. 5), the
τοῦ τοῦ τετάρτου πολλαπλασίου, τότε τὸ πρῶτον πρὸς τὸ multiple of the first (magnitude) exceeds the multiple of
δεύτερον μείζονα λόγον ἔχειν λέγεται, ἤπερ τὸ τρίτον πρὸς the second, and the multiple of the third (magnitude)
τὸ τέταρτον. does not exceed the multiple of the fourth, then the first
ηʹ. ᾿Αναλογία δὲ ἐν τρισὶν ὅροις ἐλαχίστη ἐστίν. (magnitude) is said to have a greater ratio to the second
θʹ. ῞Οταν δὲ τρία μεγέθη ἀνάλογον ᾖ, τὸ πρῶτον πρὸς than the third (magnitude has) to the fourth.
τὸ τρίτον διπλασίονα λόγον ἔχειν λέγεται ἤπερ πρὸς τὸ 8. And a proportion in three terms is the smallest
δεύτερον. (possible). $
ιʹ. ῞Οταν δὲ τέσσαρα μεγέθη ἀνάλογον ᾖ, τὸ πρῶτον 9. And when three magnitudes are proportional, the
k
πρὸς τὸ τέταρτον τριπλασίονα λόγον ἔχειν λέγεται ἤπερ first is said to have to the third the squared ratio of that
πρὸς τὸ δεύτερον, καὶ ἀεὶ ἑξῆς ὁμοίως, ὡς ἂν ἡ ἀναλογία (it has) to the second. ††
ὑπάρχῃ. 10. And when four magnitudes are (continuously)
ιαʹ. ῾Ομόλογα μεγέθη λέγεται τὰ μὲν ἡγούμενα τοῖς proportional, the first is said to have to the fourth the
ἡγουμένοις τὰ δὲ ἑπόμενα τοῖς ἑπομένοις. cubed ‡‡ ratio of that (it has) to the second. §§ And so on,
ιβʹ. ᾿Εναλλὰξ λόγος ἐστὶ λῆψις τοῦ ἡγουμένου πρὸς τὸ similarly, in successive order, whatever the (continuous)
ἡγούμενον καὶ τοῦ ἑπομένου πρὸς τὸ ἑπόμενον. proportion might be.
ιγʹ. ᾿Ανάπαλιν λόγος ἐστὶ λῆψις τοῦ ἑπομένου ὡς 11. These magnitudes are said to be corresponding
ἡγουμένου πρὸς τὸ ἡγούμενον ὡς ἑπόμενον. (magnitudes): the leading to the leading (of two ratios),
ιδʹ. Σύνθεσις λόγου ἐστὶ λῆψις τοῦ ἡγουμένου μετὰ τοῦ and the following to the following.
ἑπομένου ὡς ἑνὸς πρὸς αὐτὸ τὸ ἑπόμενον. 12. An alternate ratio is a taking of the (ratio of the)
ιεʹ. Διαίρεσις λόγου ἐστὶ λῆψις τῆς ὑπεροχῆς, ᾗ ὑπερέχει leading (magnitude) to the leading (of two equal ratios),
τὸ ἡγούμενον τοῦ ἑπομένου, πρὸς αὐτὸ τὸ ἑπόμενον. and (setting it equal to) the (ratio of the) following (mag-
ιϛʹ. ᾿Αναστροφὴ λόγου ἐστὶ λῆψις τοῦ ἡγουμένου πρὸς nitude) to the following. ¶¶
τὴν ὑπεροχήν, ᾗ ὑπερέχει τὸ ἡγούμενον τοῦ ἑπομένου. 13. An inverse ratio is a taking of the (ratio of the) fol-
ιζʹ. Δι᾿ ἴσου λόγος ἐστὶ πλειόνων ὄντων μεγεθῶν καὶ lowing (magnitude) as the leading and the leading (mag-
ἄλλων αὐτοῖς ἴσων τὸ πλῆθος σύνδυο λαμβανομένων καὶ nitude) as the following. ∗∗
ἐν τῷ αὐτῷ λόγῳ, ὅταν ᾖ ὡς ἐν τοῖς πρώτοις μεγέθεσι τὸ 14. A composition of a ratio is a taking of the (ratio of
πρῶτον πρὸς τὸ ἔσχατον, οὕτως ἐν τοῖς δευτέροις μεγέθεσι the) leading plus the following (magnitudes), as one, to
τὸ πρῶτον πρὸς τὸ ἔσχατον· ἢ ἄλλως· λῆψις τῶν ἄκρων the following (magnitude) by itself. $$


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καθ᾿ ὑπεξαίρεσιν τῶν μέσων. 15. A separation of a ratio is a taking of the (ratio
ιηʹ. Τεταραγμένη δὲ ἀναλογία ἐστίν, ὅταν τριῶν ὄντων of the) excess by which the leading (magnitude) exceeds
μεγεθῶν καὶ ἄλλων αὐτοῖς ἴσων τὸ πλῆθος γίνηται ὡς μὲν the following to the following (magnitude) by itself. kk
ἐν τοῖς πρώτοις μεγέθεσιν ἡγούμενον πρὸς ἐπόμενον, οὕτως 16. A conversion of a ratio is a taking of the (ratio
ἐν τοῖς δευτέροις μεγέθεσιν ἡγούμενον πρὸς ἑπόμενον, ὡς of the) leading (magnitude) to the excess by which the
δὲ ἐν τοῖς πρώτοις μεγέθεσιν ἑπόμενον πρὸς ἄλλο τι, οὕτως leading (magnitude) exceeds the following. †††
ἐν τοῖς δευτέροις ἄλλο τι πρὸς ἡγούμενον. 17. There being several magnitudes, and other (mag-
nitudes) of equal number to them, (which are) also in the
same ratio taken two by two, a ratio via equality (or ex
aequali) occurs when as the first is to the last in the first
(set of) magnitudes, so the first (is) to the last in the sec-
ond (set of) magnitudes. Or alternately, (it is) a taking of
the (ratio of the) outer (magnitudes) by the removal of
the inner (magnitudes). ‡‡‡
18. There being three magnitudes, and other (magni-
tudes) of equal number to them, a perturbed proportion
occurs when as the leading is to the following in the first
(set of) magnitudes, so the leading (is) to the following
in the second (set of) magnitudes, and as the following
(is) to some other (i.e., the remaining magnitude) in the
first (set of) magnitudes, so some other (is) to the leading
in the second (set of) magnitudes. §§§
† In other words, α is said to be a part of β if β = m α.
‡ In modern notation, the ratio of two magnitudes, α and β, is denoted α : β.
§ In other words, α has a ratio with respect to β if m α > β and n β > α, for some m and n.
¶ In other words, α : β :: γ : δ if and only if m α > n β whenever m γ > n δ, and m α = n β whenever m γ = n δ, and m α < n β whenever
m γ < n δ, for all m and n. This definition is the kernel of Eudoxus’ theory of proportion, and is valid even if α, β, etc., are irrational.
∗ Thus if α and β have the same ratio as γ and δ then they are proportional. In modern notation, α : β :: γ : δ.
$ In modern notation, a proportion in three terms—α, β, and γ—is written: α : β :: β : γ.
k Literally, “double”.
2
2
†† In other words, if α : β :: β : γ then α : γ :: α : β .
‡‡ Literally, “triple”.
§§ In other words, if α : β :: β : γ :: γ : δ then α : δ :: α : β .
3
3
¶¶ In other words, if α : β :: γ : δ then the alternate ratio corresponds to α : γ :: β : δ.
∗∗ In other words, if α : β then the inverse ratio corresponds to β : α.
aþ
$$ In other words, if α : β then the composed ratio corresponds to α + β : β.
kk In other words, if α : β then the separated ratio corresponds to α − β : β.
††† In other words, if α : β then the converted ratio corresponds to α : α − β.
‡‡‡ In other words, if α, β, γ are the first set of magnitudes, and δ, ǫ, ζ the second set, and α : β : γ :: δ : ǫ : ζ, then the ratio via equality (or ex
aequali) corresponds to α : γ :: δ : ζ.
§§§ In other words, if α, β, γ are the first set of magnitudes, and δ, ǫ, ζ the second set, and α : β :: δ : ǫ as well as β : γ :: ζ : δ, then the proportion
is said to be perturbed. †

Proposition 1
.
᾿Εὰν ᾖ ὁποσαοῦν μεγέθη ὁποσωνοῦν μεγεθῶν ἴσων τὸ If there are any number of magnitudes whatsoever
πλῆθος ἕκαστον ἑκάστου ἰσάκις πολλαπλάσιον, ὁσαπλάσιόν (which are) equal multiples, respectively, of some (other)
ἐστιν ἓν τῶν μεγεθῶν ἑνός, τοσαυταπλάσια ἔσται καὶ τὰ magnitudes, of equal number (to them), then as many

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πάντα τῶν πάντων. times as one of the (first) magnitudes is (divisible) by
one (of the second), so many times will all (of the first
magnitudes) also (be divisible) by all (of the second).
Α Η Β Γ Θ ∆ A G B C H D


Ε Ζ E F
῎Εστω ὁποσαοῦν μεγέθη τὰ ΑΒ, ΓΔ ὁποσωνοῦν με- Let there be any number of magnitudes whatsoever,
γεθῶν τῶν Ε, Ζ ἴσων τὸ πλῆθος ἕκαστον ἑκάστου ἰσάκις AB, CD, (which are) equal multiples, respectively, of
πολλαπλάσιον· λέγω, ὅτι ὁσαπλάσιόν ἐστι τὸ ΑΒ τοῦ Ε, some (other) magnitudes, E, F, of equal number (to
τοσαυταπλάσια ἔσται καὶ τὰ ΑΒ, ΓΔ τῶν Ε, Ζ. them). I say that as many times as AB is (divisible) by E,
᾿Επεὶ γὰρ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΑΒ τοῦ Ε καὶ so many times will AB, CD also be (divisible) by E, F.
τὸ ΓΔ τοῦ Ζ, ὅσα ἄρα ἐστὶν ἐν τῷ ΑΒ μεγέθη ἴσα τῷ Ε, For since AB, CD are equal multiples of E, F, thus
τοσαῦτα καὶ ἐν τῷ ΓΔ ἴσα τῷ Ζ. διῃρήσθω τὸ μὲν ΑΒ εἰς τὰ as many magnitudes as (there) are in AB equal to E, so
τῷ Ε μεγέθη ἴσα τὰ ΑΗ, ΗΒ, τὸ δὲ ΓΔ εἰς τὰ τῷ Ζ ἴσα τὰ many (are there) also in CD equal to F. Let AB have
ΓΘ, ΘΔ· ἔσται δὴ ἴσον τὸ πλῆθος τῶν ΑΗ, ΗΒ τῷ πλήθει been divided into magnitudes AG, GB, equal to E, and
τῶν ΓΘ, ΘΔ. καὶ ἐπεὶ ἴσον ἐστὶ τὸ μὲν ΑΗ τῷ Ε, τὸ δὲ ΓΘ CD into (magnitudes) CH, HD, equal to F. So, the
τῷ Ζ, ἴσον ἄρα τὸ ΑΗ τῷ Ε, καὶ τὰ ΑΗ, ΓΘ τοῖς Ε, Ζ. διὰ number of (divisions) AG, GB will be equal to the num-
τὰ αὐτὰ δὴ ἴσον ἐστὶ τὸ ΗΒ τῷ Ε, καὶ τὰ ΗΒ, ΘΔ τοῖς Ε, ber of (divisions) CH, HD. And since AG is equal to E,
Ζ· ὅσα ἄρα ἐστὶν ἐν τῷ ΑΒ ἴσα τῷ Ε, τοσαῦτα καὶ ἐν τοῖς and CH to F, AG (is) thus equal to E, and AG, CH to E,
ΑΒ, ΓΔ ἴσα τοῖς Ε, Ζ· ὁσαπλάσιον ἄρα ἐστὶ τὸ ΑΒ τοῦ Ε, F. So, for the same (reasons), GB is equal to E, and GB,
τοσαυταπλάσια ἔσται καὶ τὰ ΑΒ, ΓΔ τῶν Ε, Ζ. HD to E, F. Thus, as many (magnitudes) as (there) are
᾿Εὰν ἄρα ᾖ ὁποσαοῦν μεγέθη ὁποσωνοῦν μεγεθῶν in AB equal to E, so many (are there) also in AB, CD
ἴσων τὸ πλῆθος ἕκαστον ἑκάστου ἰσάκις πολλαπλάσιον, equal to E, F. Thus, as many times as AB is (divisible)
ὁσαπλάσιόν ἐστιν ἓν τῶν μεγεθῶν ἑνός, τοσαυταπλάσια by E, so many times will AB, CD also be (divisible) by
ἔσται καὶ τὰ πάντα τῶν πάντων· ὅπερ ἔδει δεῖξαι. E, F.
Thus, if there are any number of magnitudes what-
bþ many times as one of the (first) magnitudes is (divisi-
soever (which are) equal multiples, respectively, of some
(other) magnitudes, of equal number (to them), then as

ble) by one (of the second), so many times will all (of the
first magnitudes) also (be divisible) by all (of the second).
(Which is) the very thing it was required to show.

† In modern notation, this proposition reads m α + m β + · · · = m (α + β + · · · ).

†
.
Proposition 2
᾿Εὰν πρῶτον δευτέρου ἰσάκις ᾖ πολλαπλάσιον καὶ τρίτον If a first (magnitude) and a third are equal multiples
τετάρτου, ᾖ δὲ καὶ πέμπτον δευτέρου ἰσάκις πολλαπλάσιον of a second and a fourth (respectively), and a fifth (mag-
καὶ ἕκτον τετάρτου, καὶ συντεθὲν πρῶτον καὶ πέμπτον nitude) and a sixth (are) also equal multiples of the sec-
δευτέρου ἰσάκις ἔσται πολλαπλάσιον καὶ τρίτον καὶ ἕκτον ond and fourth (respectively), then the first (magnitude)
τετάρτου. and the fifth, being added together, and the third and the
Πρῶτον γὰρ τὸ ΑΒ δευτέρου τοῦ Γ ἰσάκις ἔστω πολ- sixth, (being added together), will also be equal multiples
λαπλάσιον καὶ τρίτον τὸ ΔΕ τετάρτου τοῦ Ζ, ἔστω δὲ καὶ of the second (magnitude) and the fourth (respectively).
πέμπτον τὸ ΒΗ δευτέρου τοῦ Γ ἰσάκις πολλαπλάσιον καὶ For let a first (magnitude) AB and a third DE be
ἕκτον τὸ ΕΘ τετάρτου τοῦ Ζ· λέγω, ὅτι καὶ συντεθὲν equal multiples of a second C and a fourth F (respec-
πρῶτον καὶ πέμπτον τὸ ΑΗ δευτέρου τοῦ Γ ἰσάκις ἔσται tively). And let a fifth (magnitude) BG and a sixth EH
πολλαπλάσιον καὶ τρίτον καὶ ἕκτον τὸ ΔΘ τετάρτου τοῦ Ζ. also be (other) equal multiples of the second C and the
fourth F (respectively). I say that the first (magnitude)
and the fifth, being added together, (to give) AG, and the
third (magnitude) and the sixth, (being added together,

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to give) DH, will also be equal multiples of the second
(magnitude) C and the fourth F (respectively).
Α Β Η A B G


Γ C

∆ Ε Θ D E H



Ζ F
᾿Επεὶ γὰρ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΑΒ τοῦ Γ καὶ τὸ For since AB and DE are equal multiples of C and F
ΔΕ τοῦ Ζ, ὅσα ἄρα ἐστὶν ἐν τῷ ΑΒ ἴσα τῷ Γ, τοσαῦτα καὶ (respectively), thus as many (magnitudes) as (there) are
ἐν τῷ ΔΕ ἴσα τῷ Ζ. διὰ τὰ αὐτὰ δὴ καὶ ὅσα ἐστὶν ἐν τῷ ΒΗ in AB equal to C, so many (are there) also in DE equal to
ἴσα τῷ Γ, τοσαῦτα καὶ ἐν τῷ ΕΘ ἴσα τῷ Ζ· ὅσα ἄρα ἐστὶν ἐν F. And so, for the same (reasons), as many (magnitudes)
ὅλῳ τῷ ΑΗ ἴσα τῷ Γ, τοσαῦτα καὶ ἐν ὅλῳ τῷ ΔΘ ἴσα τῷ Ζ· as (there) are in BG equal to C, so many (are there)
ὁσαπλάσιον ἄρα ἐστὶ τὸ ΑΗ τοῦ Γ, τοσαυταπλάσιον ἔσται also in EH equal to F. Thus, as many (magnitudes) as
καὶ τὸ ΔΘ τοῦ Ζ. καὶ συντεθὲν ἄρα πρῶτον καὶ πέμπτον τὸ (there) are in the whole of AG equal to C, so many (are
ΑΗ δευτέρου τοῦ Γ ἰσάκις ἔσται πολλαπλάσιον καὶ τρίτον there) also in the whole of DH equal to F. Thus, as many
καὶ ἕκτον τὸ ΔΘ τετάρτου τοῦ Ζ. times as AG is (divisible) by C, so many times will DH
᾿Εὰν ἄρα πρῶτον δευτέρου ἰσάκις ᾖ πολλαπλάσιον καὶ also be divisible by F. Thus, the first (magnitude) and
τρίτον τετάρτου, ᾖ δὲ καὶ πέμπτον δευτέρου ἰσάκις πολ- the fifth, being added together, (to give) AG, and the
λαπλάσιον καὶ ἕκτον τετάρτου, καὶ συντεθὲν πρῶτον καὶ third (magnitude) and the sixth, (being added together,
πέμπτον δευτέρου ἰσάκις ἔσται πολλαπλάσιον καὶ τρίτον καὶ to give) DH, will also be equal multiples of the second
ἕκτον τετάρτου· ὅπερ ἔδει δεῖξαι. (magnitude) C and the fourth F (respectively).
Thus, if a first (magnitude) and a third are equal mul-
tiples of a second and a fourth (respectively), and a fifth
gþ tude) and the fifth, being added together, and the third
(magnitude) and a sixth (are) also equal multiples of the
second and fourth (respectively), then the first (magni-

and sixth, (being added together), will also be equal mul-
tiples of the second (magnitude) and the fourth (respec-
tively). (Which is) the very thing it was required to show.

† In modern notation, this propostion reads m α + n α = (m + n) α.

.
†
Proposition 3
᾿Εὰν πρῶτον δευτέρου ἰσάκις ᾖ πολλαπλάσιον καὶ τρίτον If a first (magnitude) and a third are equal multiples
τετάρτου, ληφθῇ δὲ ἰσάκις πολλαπλάσια τοῦ τε πρώτου of a second and a fourth (respectively), and equal multi-
καὶ τρίτου, καὶ δι᾿ ἴσου τῶν ληφθέντων ἑκάτερον ἑκατέρου ples are taken of the first and the third, then, via equality,
ἰσάκις ἔσται πολλαπλάσιον τὸ μὲν τοῦ δευτέρου τὸ δὲ τοῦ the (magnitudes) taken will also be equal multiples of the
τετάρτου. second (magnitude) and the fourth, respectively.
Πρῶτον γὰρ τὸ Α δευτέρου τοῦ Β ἰσάκις ἔστω πολ- For let a first (magnitude) A and a third C be equal
λαπλάσιον καὶ τρίτον τὸ Γ τετάρτου τοῦ Δ, καὶ εἰλήφθω multiples of a second B and a fourth D (respectively),
τῶν Α, Γ ἰσάκις πολλαπλάσια τὰ ΕΖ, ΗΘ· λέγω, ὅτι ἰσάκις and let the equal multiples EF and GH have been taken
ἐστὶ πολλαπλάσιον τὸ ΕΖ τοῦ Β καὶ τὸ ΗΘ τοῦ Δ. of A and C (respectively). I say that EF and GH are
᾿Επεὶ γὰρ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΕΖ τοῦ Α καὶ equal multiples of B and D (respectively).
τὸ ΗΘ τοῦ Γ, ὅσα ἄρα ἐστὶν ἐν τῷ ΕΖ ἴσα τῷ Α, τοσαῦτα For since EF and GH are equal multiples of A and
καὶ ἐν τῷ ΗΘ ἴσα τῷ Γ. διῃρήσθω τὸ μὲν ΕΖ εἰς τὰ τῷ Α C (respectively), thus as many (magnitudes) as (there)
μεγέθη ἴσα τὰ ΕΚ, ΚΖ, τὸ δὲ ΗΘ εἰς τὰ τῷ Γ ἴσα τὰ ΗΛ, are in EF equal to A, so many (are there) also in GH


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ΛΘ· ἔσται δὴ ἴσον τὸ πλῆθος τῶν ΕΚ, ΚΖ τῷ πλήθει τῶν equal to C. Let EF have been divided into magnitudes
ΗΛ, ΛΘ. καὶ ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ Α τοῦ Β καὶ EK, KF equal to A, and GH into (magnitudes) GL, LH
τὸ Γ τοῦ Δ, ἴσον δὲ τὸ μὲν ΕΚ τῷ Α, τὸ δὲ ΗΛ τῷ Γ, equal to C. So, the number of (magnitudes) EK, KF
ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΕΚ τοῦ Β καὶ τὸ ΗΛ τοῦ will be equal to the number of (magnitudes) GL, LH.
Δ. διὰ τὰ αὐτὰ δὴ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΚΖ τοῦ Β And since A and C are equal multiples of B and D (re-
καὶ τὸ ΛΘ τοῦ Δ. ἐπεὶ οὖν πρῶτον τὸ ΕΚ δευτέρου τοῦ Β spectively), and EK (is) equal to A, and GL to C, EK
ἴσάκις ἐστὶ πολλαπλάσιον καὶ τρίτον τὸ ΗΛ τετάρτου τοῦ and GL are thus equal multiples of B and D (respec-
Δ, ἔστι δὲ καὶ πέμπτον τὸ ΚΖ δευτέρου τοῦ Β ἰσάκις πολ- tively). So, for the same (reasons), KF and LH are equal
λαπλάσιον καὶ ἕκτον τὸ ΛΘ τετάρτου τοῦ Δ, καὶ συντεθὲν multiples of B and D (respectively). Therefore, since the
ἄρα πρῶτον καὶ πέμπτον τὸ ΕΖ δευτέρου τοῦ Β ἰσάκις ἐστὶ first (magnitude) EK and the third GL are equal mul-
πολλαπλάσιον καὶ τρίτον καὶ ἕκτον τὸ ΗΘ τετάρτου τοῦ Δ. tiples of the second B and the fourth D (respectively),
and the fifth (magnitude) KF and the sixth LH are also
equal multiples of the second B and the fourth D (re-
spectively), then the first (magnitude) and fifth, being
added together, (to give) EF, and the third (magnitude)
and sixth, (being added together, to give) GH, are thus
also equal multiples of the second (magnitude) B and the
fourth D (respectively) [Prop. 5.2].
Α A

Β B

Ε Κ Ζ E K F


Γ C

∆ D

Η Λ Θ G L H
dþ tively. (Which is) the very thing it was required to show.
᾿Εὰν ἄρα πρῶτον δευτέρου ἰσάκις ᾖ πολλαπλάσιον Thus, if a first (magnitude) and a third are equal mul-
καὶ τρίτον τετάρτου, ληφθῇ δὲ τοῦ πρώτου καὶ τρίτου tiples of a second and a fourth (respectively), and equal
ἰσάκις πολλαπλάσια, καὶ δι᾿ ἴσου τῶν ληφθέντων ἑκάτερον multiples are taken of the first and the third, then, via
ἑκατέρου ἰσάκις ἔσται πολλαπλάσιον τὸ μὲν τοῦ δευτέρου equality, the (magnitudes) taken will also be equal mul-
τὸ δὲ τοῦ τετάρτου· ὅπερ ἔδει δεῖξαι. tiples of the second (magnitude) and the fourth, respec-


† In modern notation, this proposition reads m(n α) = (m n) α.

.
Proposition 4
†
᾿Εὰν πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἔχῃ λόγον καὶ If a first (magnitude) has the same ratio to a second
τρίτον πρὸς τέταρτον, καὶ τὰ ἰσάκις πολλαπλάσια τοῦ τε that a third (has) to a fourth then equal multiples of the
πρώτου καὶ τρίτου πρὸς τὰ ἰσάκις πολλαπλάσια τοῦ δευτέρου first (magnitude) and the third will also have the same
καὶ τετάρτου καθ᾿ ὁποιονοῦν πολλαπλασιασμὸν τὸν αὐτὸν ratio to equal multiples of the second and the fourth, be-
ἕξει λόγον ληφθέντα κατάλληλα. ing taken in corresponding order, according to any kind
Πρῶτον γὰρ τὸ Α πρὸς δεύτερον τὸ Β τὸν αὐτὸν ἐχέτω of multiplication whatsoever.
λόγον καὶ τρίτον τὸ Γ πρὸς τέταρτον τὸ Δ, καὶ εἰλήφθω For let a first (magnitude) A have the same ratio to
τῶν μὲν Α, Γ ἰσάκις πολλαπλάσια τὰ Ε, Ζ, τῶν δὲ Β, Δ a second B that a third C (has) to a fourth D. And let
ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Η, Θ· λέγω, ὅτι equal multiples E and F have been taken of A and C
ἐστὶν ὡς τὸ Ε πρὸς τὸ Η, οὕτως τὸ Ζ πρὸς τὸ Θ. (respectively), and other random equal multiples G and


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H of B and D (respectively). I say that as E (is) to G, so
F (is) to H.
Α A
Β B
Ε E

Η G
Κ K
Μ M

Γ C
∆ D
Ζ F

Θ H
Λ L
Ν N
Εἰλήφθω γὰρ τῶν μὲν Ε, Ζ ἰσάκις πολλαπλάσια τὰ Κ, For let equal multiples K and L have been taken of E
Λ, τῶν δὲ Η, Θ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Μ, and F (respectively), and other random equal multiples
Ν. M and N of G and H (respectively).
[Καὶ] ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ μὲν Ε τοῦ Α, τὸ [And] since E and F are equal multiples of A and
δὲ Ζ τοῦ Γ, καὶ εἴληπται τῶν Ε, Ζ ἴσάκις πολλαπλάσια τὰ Κ, C (respectively), and the equal multiples K and L have
Λ, ἴσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ Κ τοῦ Α καὶ τὸ Λ τοῦ been taken of E and F (respectively), K and L are thus
Γ. διὰ τὰ αὐτὰ δὴ ἰσάκις ἐστὶ πολλαπλάσιον τὸ Μ τοῦ Β καὶ equal multiples of A and C (respectively) [Prop. 5.3]. So,
τὸ Ν τοῦ Δ. καὶ ἐπεί ἐστιν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ for the same (reasons), M and N are equal multiples of
πρὸς τὸ Δ, καὶ εἴληπται τῶν μὲν Α, Γ ἰσάκις πολλαπλάσια B and D (respectively). And since as A is to B, so C (is)
τὰ Κ, Λ, τῶν δὲ Β, Δ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια to D, and the equal multiples K and L have been taken
τὰ Μ, Ν, εἰ ἄρα ὑπερέχει τὸ Κ τοῦ Μ, ὑπερέχει καὶ τὸ Λ of A and C (respectively), and the other random equal
τοῦ Ν, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον. καί ἐστι multiples M and N of B and D (respectively), then if K
τὰ μὲν Κ, Λ τῶν Ε, Ζ ἰσάκις πολλαπλάσια, τὰ δὲ Μ, Ν τῶν exceeds M then L also exceeds N, and if (K is) equal (to
Η, Θ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια· ἔστιν ἄρα ὡς τὸ M then L is also) equal (to N), and if (K is) less (than M
Ε πρὸς τὸ Η, οὕτως τὸ Ζ πρὸς τὸ Θ. then L is also) less (than N) [Def. 5.5]. And K and L are
᾿Εὰν ἄρα πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἔχῃ λόγον equal multiples of E and F (respectively), and M and N
καὶ τρίτον πρὸς τέταρτον, καὶ τὰ ἰσάκις πολλαπλάσια τοῦ τε other random equal multiples of G and H (respectively).
πρώτου καὶ τρίτου πρὸς τὰ ἰσάκις πολλαπλάσια τοῦ δευτέρου Thus, as E (is) to G, so F (is) to H [Def. 5.5].
καὶ τετάρτου τὸν αὐτὸν ἕξει λόγον καθ᾿ ὁποιονοῦν πολλα- Thus, if a first (magnitude) has the same ratio to a
πλασιασμὸν ληφθέντα κατάλληλα· ὅπερ ἔδει δεῖξαι. second that a third (has) to a fourth then equal multi-
ples of the first (magnitude) and the third will also have
the same ratio to equal multiples of the second and the
fourth, being taken in corresponding order, according to
any kind of multiplication whatsoever. (Which is) the
very thing it was required to show.

† In modern notation, this proposition reads that if α : β :: γ : δ then m α : n β :: m γ : n δ, for all m and n.







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†
Proposition 5
.
᾿Εὰν μέγεθος μεγέθους ἰσάκις ᾖ πολλαπλάσιον, ὅπερ If a magnitude is the same multiple of a magnitude
ἀφαιρεθὲν ἀφαιρεθέντος, καὶ τὸ λοιπὸν τοῦ λοιποῦ ἰσάκις that a (part) taken away (is) of a (part) taken away (re-
ἔσται πολλαπλάσιον, ὁσαπλάσιόν ἐστι τὸ ὅλον τοῦ ὅλου. spectively) then the remainder will also be the same mul-
tiple of the remainder as that which the whole (is) of the
whole (respectively).
Α Ε Β A E B

Η Γ Ζ ∆ G C F D

Μέγεθος γὰρ τὸ ΑΒ μεγέθους τοῦ ΓΔ ἰσάκις ἔστω πολ- For let the magnitude AB be the same multiple of the
λαπλάσιον, ὅπερ ἀφαιρεθὲν τὸ ΑΕ ἀφαιρεθέντος τοῦ ΓΖ· magnitude CD that the (part) taken away AE (is) of the
λέγω, ὅτι καὶ λοιπὸν τὸ ΕΒ λοιποῦ τοῦ ΖΔ ἰσάκις ἔσται (part) taken away CF (respectively). I say that the re-
πολλαπλάσιον, ὁσαπλάσιόν ἐστιν ὅλον τὸ ΑΒ ὅλου τοῦ ΓΔ. mainder EB will also be the same multiple of the remain-
῾Οσαπλάσιον γάρ ἐστι τὸ ΑΕ τοῦ ΓΖ, τοσαυταπλάσιον der FD as that which the whole AB (is) of the whole CD
γεγονέτω καὶ τὸ ΕΒ τοῦ ΓΗ. (respectively).
Καὶ ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΑΕ τοῦ ΓΖ καὶ τὸ For as many times as AE is (divisible) by CF, so many
ΕΒ τοῦ ΗΓ, ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΑΕ τοῦ ΓΖ times let EB also have been made (divisible) by CG.
καὶ τὸ ΑΒ τοῦ ΗΖ. κεῖται δὲ ἰσάκις πολλαπλάσιον τὸ ΑΕ And since AE and EB are equal multiples of CF and
τοῦ ΓΖ καὶ τὸ ΑΒ τοῦ ΓΔ. ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ GC (respectively), AE and AB are thus equal multiples
ΑΒ ἑκατέρου τῶν ΗΖ, ΓΔ· ἴσον ἄρα τὸ ΗΖ τῷ ΓΔ. κοινὸν of CF and GF (respectively) [Prop. 5.1]. And AE and
ἀφῃρήσθω τὸ ΓΖ· λοιπὸν ἄρα τὸ ΗΓ λοιπῷ τῷ ΖΔ ἴσον AB are assumed (to be) equal multiples of CF and CD
ἐστίν. καὶ ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΑΕ τοῦ ΓΖ (respectively). Thus, AB is an equal multiple of each
καὶ τὸ ΕΒ τοῦ ΗΓ, ἴσον δὲ τὸ ΗΓ τῷ ΔΖ, ἰσάκις ἄρα ἐστὶ of GF and CD. Thus, GF (is) equal to CD. Let CF
πολλαπλάσιον τὸ ΑΕ τοῦ ΓΖ καὶ τὸ ΕΒ τοῦ ΖΔ. ἰσάκις δὲ have been subtracted from both. Thus, the remainder
ὑπόκειται πολλαπλάσιον τὸ ΑΕ τοῦ ΓΖ καὶ τὸ ΑΒ τοῦ ΓΔ· GC is equal to the remainder FD. And since AE and
ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΕΒ τοῦ ΖΔ καὶ τὸ ΑΒ EB are equal multiples of CF and GC (respectively),
τοῦ ΓΔ. καὶ λοιπὸν ἄρα τὸ ΕΒ λοιποῦ τοῦ ΖΔ ἰσάκις ἔσται and GC (is) equal to DF, AE and EB are thus equal
πολλαπλάσιον, ὁσαπλάσιόν ἐστιν ὅλον τὸ ΑΒ ὅλου τοῦ ΓΔ. multiples of CF and FD (respectively). And AE and
᾿Εὰν ἄρα μέγεθος μεγέθους ἰσάκις ᾖ πολλαπλάσιον, AB are assumed (to be) equal multiples of CF and CD
ὅπερ ἀφαιρεθὲν ἀφαιρεθέντος, καὶ τὸ λοιπὸν τοῦ λοιποῦ (respectively). Thus, EB and AB are equal multiples of
ἰσάκις ἔσται πολλαπλάσιον, ὁσαπλάσιόν ἐστι καὶ τὸ ὅλον FD and CD (respectively). Thus, the remainder EB will
τοῦ ὅλου· ὅπερ ἔδει δεῖξαι. also be the same multiple of the remainder FD as that
which the whole AB (is) of the whole CD (respectively).
þ (respectively) then the remainder will also be the same
Thus, if a magnitude is the same multiple of a magni-
tude that a (part) taken away (is) of a (part) taken away

multiple of the remainder as that which the whole (is) of
the whole (respectively). (Which is) the very thing it was
required to show.
† In modern notation, this proposition reads m α − m β = m (α − β).

†
Proposition 6
.
᾿Εὰν δύο μεγέθη δύο μεγεθῶν ἰσάκις ᾖ πολλαπλάσια, If two magnitudes are equal multiples of two (other)
καὶ ἀφαιρεθέντα τινὰ τῶν αὐτῶν ἰσάκις ᾖ πολλαπλάσια, καὶ magnitudes, and some (parts) taken away (from the for-
τὰ λοιπὰ τοῖς αὐτοῖς ἤτοι ἴσα ἐστὶν ἢ ἰσάκις αὐτῶν πολ- mer magnitudes) are equal multiples of the latter (mag-
λαπλάσια. nitudes, respectively), then the remainders are also either
Δύο γὰρ μεγέθη τὰ ΑΒ, ΓΔ δύο μεγεθῶν τῶν Ε, Ζ equal to the latter (magnitudes), or (are) equal multiples


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ἰσάκις ἔστω πολλαπλάσια, καὶ ἀφαιρεθέντα τὰ ΑΗ, ΓΘ τῶν of them (respectively). ELEMENTS BOOK 5
αὐτῶν τῶν Ε, Ζ ἰσάκις ἔστω πολλαπλάσια· λέγω, ὅτι καὶ For let two magnitudes AB and CD be equal multi-
λοιπὰ τὰ ΗΒ, ΘΔ τοῖς Ε, Ζ ἤτοι ἴσα ἐστὶν ἢ ἰσάκις αὐτῶν ples of two magnitudes E and F (respectively). And let
πολλαπλάσια. the (parts) taken away (from the former) AG and CH be
equal multiples of E and F (respectively). I say that the
remainders GB and HD are also either equal to E and F
(respectively), or (are) equal multiples of them.
Α Η Β A G B




Ε E

Κ Γ Θ ∆ K C H D




Ζ F
῎Εστω γὰρ πρότερον τὸ ΗΒ τῷ Ε ἴσον· λέγω, ὅτι καὶ For let GB be, first of all, equal to E. I say that HD is
τὸ ΘΔ τῷ Ζ ἴσον ἐστίν. also equal to F.
Κείσθω γὰρ τῷ Ζ ἴσον τὸ ΓΚ. ἐπεὶ ἰσάκις ἐστὶ πολ- For let CK be made equal to F. Since AG and CH
λαπλάσιον τὸ ΑΗ τοῦ Ε καὶ τὸ ΓΘ τοῦ Ζ, ἴσον δὲ τὸ μὲν ΗΒ are equal multiples of E and F (respectively), and GB
τῷ Ε, τὸ δὲ ΚΓ τῷ Ζ, ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΑΒ (is) equal to E, and KC to F, AB and KH are thus equal
τοῦ Ε καὶ τὸ ΚΘ τοῦ Ζ. ἰσάκις δὲ ὑπόκειται πολλαπλάσιον multiples of E and F (respectively) [Prop. 5.2]. And AB
τὸ ΑΒ τοῦ Ε καὶ τὸ ΓΔ τοῦ Ζ· ἴσάκις ἄρα ἐστὶ πολλαπλάσιον and CD are assumed (to be) equal multiples of E and F
τὸ ΚΘ τοῦ Ζ καὶ τὸ ΓΔ τοῦ Ζ. ἐπεὶ οὖν ἑκάτερον τῶν ΚΘ, (respectively). Thus, KH and CD are equal multiples of
ΓΔ τοῦ Ζ ἰσάκις ἐστὶ πολλαπλάσιον, ἴσον ἄρα ἐστὶ τὸ ΚΘ F and F (respectively). Therefore, KH and CD are each
τῷ ΓΔ. κοινὸν ἀφῃρήσθω τὸ ΓΘ· λοιπὸν ἄρα τὸ ΚΓ λοιπῷ equal multiples of F. Thus, KH is equal to CD. Let CH
τῷ ΘΔ ἴσον ἐστίν. ἀλλὰ τὸ Ζ τῷ ΚΓ ἐστιν ἴσον· καὶ τὸ have be taken away from both. Thus, the remainder KC
ΘΔ ἄρα τῷ Ζ ἴσον ἐστίν. ὥστε εἰ τὸ ΗΒ τῷ Ε ἴσον ἐστίν, is equal to the remainder HD. But, F is equal to KC.
καὶ τὸ ΘΔ ἴσον ἔσται τῷ Ζ. Thus, HD is also equal to F. Hence, if GB is equal to E
῾Ομοίως δὴ δείξομεν, ὅτι, κᾂν πολλαπλάσιον ᾖ τὸ ΗΒ then HD will also be equal to F.
τοῦ Ε, τοσαυταπλάσιον ἔσται καὶ τὸ ΘΔ τοῦ Ζ. So, similarly, we can show that even if GB is a multi-
᾿Εὰν ἄρα δύο μεγέθη δύο μεγεθῶν ἰσάκις ᾖ πολ- ple of E then HD will also be the same multiple of F.
λαπλάσια, καὶ ἀφαιρεθέντα τινὰ τῶν αὐτῶν ἰσάκις ᾖ πολ- Thus, if two magnitudes are equal multiples of two
zþ either equal to the latter (magnitudes), or (are) equal
λαπλάσια, καὶ τὰ λοιπὰ τοῖς αὐτοῖς ἤτοι ἴσα ἐστὶν ἢ ἰσάκις (other) magnitudes, and some (parts) taken away (from
αὐτῶν πολλαπλάσια· ὅπερ ἔδει δεῖξαι. the former magnitudes) are equal multiples of the latter
(magnitudes, respectively), then the remainders are also

multiples of them (respectively). (Which is) the very
thing it was required to show.
† In modern notation, this proposition reads m α − n α = (m − n) α.

Proposition 7
.
Τὰ ἴσα πρὸς τὸ αὐτὸ τὸν αὐτὸν ἔχει λόγον καὶ τὸ αὐτὸ Equal (magnitudes) have the same ratio to the same
πρὸς τὰ ἴσα. (magnitude), and the latter (magnitude has the same ra-
῎Εστω ἴσα μεγέθη τὰ Α, Β, ἄλλο δέ τι, ὃ ἔτυχεν, tio) to the equal (magnitudes).
μέγεθος τὸ Γ· λέγω, ὅτι ἑκάτερον τῶν Α, Β πρὸς τὸ Γ Let A and B be equal magnitudes, and C some other
τὸν αὐτὸν ἔχει λόγον, καὶ τὸ Γ πρὸς ἑκάτερον τῶν Α, Β. random magnitude. I say that A and B each have the


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same ratio to C, and (that) C (has the same ratio) to
each of A and B.
Α ∆ A D
Β Ε B E

Γ Ζ C F
Εἰλήφθω γὰρ τῶν μὲν Α, Β ἰσάκις πολλαπλάσια τὰ Δ, For let the equal multiples D and E have been taken
Ε, τοῦ δὲ Γ ἄλλο, ὃ ἔτυχεν, πολλαπλάσιον τὸ Ζ. of A and B (respectively), and the other random multiple
᾿Επεὶ οὖν ἰσάκις ἐστὶ πολλαπλάσιον τὸ Δ τοῦ Α καὶ τὸ F of C.
Ε τοῦ Β, ἴσον δὲ τὸ Α τῷ Β, ἴσον ἄρα καὶ τὸ Δ τῷ Ε. ἄλλο Therefore, since D and E are equal multiples of A
δέ, ὅ ἔτυχεν, τὸ Ζ. Εἰ ἄρα ὑπερέχει τὸ Δ τοῦ Ζ, ὑπερέχει and B (respectively), and A (is) equal to B, D (is) thus
καὶ τὸ Ε τοῦ Ζ, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον. also equal to E. And F (is) different, at random. Thus, if
καί ἐστι τὰ μὲν Δ, Ε τῶν Α, Β ἰσάκις πολλαπλάσια, τὸ δὲ D exceeds F then E also exceeds F, and if (D is) equal
Ζ τοῦ Γ ἄλλο, ὃ ἔτυχεν, πολλαπλάσιον· ἔστιν ἄρα ὡς τὸ Α (to F then E is also) equal (to F), and if (D is) less
πρὸς τὸ Γ, οὕτως τὸ Β πρὸς τὸ Γ. (than F then E is also) less (than F). And D and E are
Λέγω [δή], ὅτι καὶ τὸ Γ πρὸς ἑκάτερον τῶν Α, Β τὸν equal multiples of A and B (respectively), and F another
αὐτὸν ἔχει λόγον. random multiple of C. Thus, as A (is) to C, so B (is) to
Τῶν γὰρ αὐτῶν κατασκευασθέντων ὁμοίως δείξομεν, C [Def. 5.5].
†
ὅτι ἴσον ἐστὶ τὸ Δ τῷ Ε· ἄλλο δέ τι τὸ Ζ· εἰ ἄρα ὑπερέχει [So] I say that C also has the same ratio to each of A
τὸ Ζ τοῦ Δ, ὑπερέχει καὶ τοῦ Ε, καὶ εἰ ἴσον, ἴσον, καὶ εἰ and B.
ἔλαττον, ἔλαττον. καί ἐστι τὸ μὲν Ζ τοῦ Γ πολλαπλάσιον, For, similarly, we can show, by the same construction,
τὰ δὲ Δ, Ε τῶν Α, Β ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια· that D is equal to E. And F (has) some other (value).
ìri sma and B. Thus, as C (is) to A, so C (is) to B [Def. 5.5].
ἔστιν ἄρα ὡς τὸ Γ πρὸς τὸ Α, οὕτως τὸ Γ πρὸς τὸ Β. Thus, if F exceeds D then it also exceeds E, and if (F is)
Τὰ ἴσα ἄρα πρὸς τὸ αὐτὸ τὸν αὐτὸν ἔχει λόγον καὶ τὸ equal (to D then it is also) equal (to E), and if (F is) less
αὐτὸ πρὸς τὰ ἴσα. (than D then it is also) less (than E). And F is a multiple
of C, and D and E other random equal multiples of A

Thus, equal (magnitudes) have the same ratio to the
same (magnitude), and the latter (magnitude has the
same ratio) to the equal (magnitudes).
hþ (Which is) the very thing it was required to show.
.
Corollary
‡
᾿Εκ δὴ τούτου φανερόν, ὅτι ἐὰν μεγέθη τινὰ ἀνάλογον So (it is) clear, from this, that if some magnitudes are
ᾖ, καὶ ἀνάπαλιν ἀνάλογον ἔσται. ὅπερ ἔδει δεῖξαι. proportional then they will also be proportional inversely.


† The Greek text has “E”, which is obviously a mistake.
‡ In modern notation, this corollary reads that if α : β :: γ : δ then β : α :: δ : γ.

Proposition 8
.
Τῶν ἀνίσων μεγεθῶν τὸ μεῖζον πρὸς τὸ αὐτὸ μείζονα For unequal magnitudes, the greater (magnitude) has
λόγον ἔχει ἤπερ τὸ ἔλαττον. καὶ τὸ αὐτὸ πρὸς τὸ ἔλαττον a greater ratio than the lesser to the same (magnitude).
μείζονα λόγον ἔχει ἤπερ πρὸς τὸ μεῖζον. And the latter (magnitude) has a greater ratio to the
῎Εστω ἄνισα μεγέθη τὰ ΑΒ, Γ, καὶ ἔστω μεῖζον τὸ ΑΒ, lesser (magnitude) than to the greater.
ἄλλο δέ, ὃ ἔτυχεν, τὸ Δ· λέγω, ὅτι τὸ ΑΒ πρὸς τὸ Δ Let AB and C be unequal magnitudes, and let AB be
μείζονα λόγον ἔχει ἤπερ τὸ Γ πρὸς τὸ Δ, καὶ τὸ Δ πρὸς the greater (of the two), and D another random magni-
τὸ Γ μείζονα λόγον ἔχει ἤπερ πρὸς τὸ ΑΒ. tude. I say that AB has a greater ratio to D than C (has)
to D, and (that) D has a greater ratio to C than (it has)
to AB.


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A E B

G
G E A B Z H A E B ELEMENTS BOOK 5
H Z
A
E B
C F G H C F G H
D
D


K
K


D
D


L
M
M
N
N
᾿Επεὶ γὰρ μεῖζόν ἐστι τὸ ΑΒ τοῦ Γ, κείσθω τῷ Γ ἴσον L For since AB is greater than C, let BE be made equal
τὸ ΒΕ· τὸ δὴ ἔλασσον τῶν ΑΕ, ΕΒ πολλαπλασιαζόμενον to C. So, the lesser of AE and EB, being multiplied, will
ἔσται ποτὲ τοῦ Δ μεῖζον. ἔστω πρότερον τὸ ΑΕ ἔλαττον sometimes be greater than D [Def. 5.4]. First of all, let
τοῦ ΕΒ, καὶ πεπολλαπλασιάσθω τὸ ΑΕ, καὶ ἔστω αὐτοῦ AE be less than EB, and let AE have been multiplied,
πολλαπλάσιον τὸ ΖΗ μεῖζον ὂν τοῦ Δ, καὶ ὁσαπλάσιόν ἐστι and let FG be a multiple of it which (is) greater than
τὸ ΖΗ τοῦ ΑΕ, τοσαυταπλάσιον γεγονέτω καὶ τὸ μὲν ΗΘ D. And as many times as FG is (divisible) by AE, so
τοῦ ΕΒ τὸ δὲ Κ τοῦ Γ· καὶ εἰλήφθω τοῦ Δ διπλάσιον μὲν many times let GH also have become (divisible) by EB,
τὸ Λ, τριπλάσιον δὲ τὸ Μ, καὶ ἑξῆς ἑνὶ πλεῖον, ἕως ἂν τὸ and K by C. And let the double multiple L of D have
λαμβανόμενον πολλαπλάσιον μὲν γένηται τοῦ Δ, πρώτως δὲ been taken, and the triple multiple M, and several more,
μεῖζον τοῦ Κ. εἰλήφθω, καὶ ἔστω τὸ Ν τετραπλάσιον μὲν (each increasing) in order by one, until the (multiple)
τοῦ Δ, πρώτως δὲ μεῖζον τοῦ Κ. taken becomes the first multiple of D (which is) greater
᾿Επεὶ οὖν τὸ Κ τοῦ Ν πρώτως ἐστὶν ἔλαττον, τὸ Κ ἄρα than K. Let it have been taken, and let it also be the
τοῦ Μ οὔκ ἐστιν ἔλαττον. καὶ ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον quadruple multiple N of D—the first (multiple) greater
τὸ ΖΗ τοῦ ΑΕ καὶ τὸ ΗΘ τοῦ ΕΒ, ἰσάκις ἄρα ἐστὶ πολ- than K.
λαπλάσιον τὸ ΖΗ τοῦ ΑΕ καὶ τὸ ΖΘ τοῦ ΑΒ. ἰσάκις δέ Therefore, since K is less than N first, K is thus not
ἐστι πολλαπλάσιον τὸ ΖΗ τοῦ ΑΕ καὶ τὸ Κ τοῦ Γ· ἰσάκις less than M. And since FG and GH are equal multi-
ἄρα ἐστὶ πολλαπλάσιον τὸ ΖΘ τοῦ ΑΒ καὶ τὸ Κ τοῦ Γ. τὰ ples of AE and EB (respectively), FG and FH are thus
ΖΘ, Κ ἄρα τῶν ΑΒ, Γ ἰσάκις ἐστὶ πολλαπλάσια. πάλιν, ἐπεὶ equal multiples of AE and AB (respectively) [Prop. 5.1].
ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΗΘ τοῦ ΕΒ καὶ τὸ Κ τοῦ Γ, And FG and K are equal multiples of AE and C (re-
ἴσον δὲ τὸ ΕΒ τῷ Γ, ἴσον ἄρα καὶ τὸ ΗΘ τῷ Κ. τὸ δὲ Κ spectively). Thus, FH and K are equal multiples of AB
τοῦ Μ οὔκ ἐστιν ἔλαττον· οὐδ᾿ ἄρα τὸ ΗΘ τοῦ Μ ἔλαττόν and C (respectively). Thus, FH, K are equal multiples
ἐστιν. μεῖζον δὲ τὸ ΖΗ τοῦ Δ· ὅλον ἄρα τὸ ΖΘ συναμ- of AB, C. Again, since GH and K are equal multiples
φοτέρων τῶν Δ, Μ μεῖζόν ἐστιν. ἀλλὰ συναμφότερα τὰ Δ, of EB and C, and EB (is) equal to C, GH (is) thus also
Μ τῷ Ν ἐστιν ἴσα, ἐπειδήπερ τὸ Μ τοῦ Δ τριπλάσιόν ἐστιν, equal to K. And K is not less than M. Thus, GH not less
συναμφότερα δὲ τὰ Μ, Δ τοῦ Δ ἐστι τετραπλάσια, ἔστι δὲ than M either. And FG (is) greater than D. Thus, the
καὶ τὸ Ν τοῦ Δ τετραπλάσιον· συναμφότερα ἄρα τὰ Μ, Δ whole of FH is greater than D and M (added) together.
τῷ Ν ἴσα ἐστίν. ἀλλὰ τὸ ΖΘ τῶν Μ, Δ μεῖζόν ἐστιν· τὸ But, D and M (added) together is equal to N, inasmuch
ΖΘ ἄρα τοῦ Ν ὑπερέχει· τὸ δὲ Κ τοῦ Ν οὐχ ὑπερέχει. καί as M is three times D, and M and D (added) together is
ἐστι τὰ μὲν ΖΘ, Κ τῶν ΑΒ, Γ ἰσάκις πολλαπλάσια, τὸ δὲ Ν four times D, and N is also four times D. Thus, M and D
τοῦ Δ ἄλλο, ὃ ἔτυχεν, πολλαπλάσιον· τὸ ΑΒ ἄρα πρὸς τὸ (added) together is equal to N. But, FH is greater than
Δ μείζονα λόγον ἔχει ἤπερ τὸ Γ πρὸς τὸ Δ. M and D. Thus, FH exceeds N. And K does not exceed
Λέγω δή, ὅτι καὶ τὸ Δ πρὸς τὸ Γ μείζονα λόγον ἔχει N. And FH, K are equal multiples of AB, C, and N
ἤπερ τὸ Δ πρὸς τὸ ΑΒ. another random multiple of D. Thus, AB has a greater
Τῶν γὰρ αὐτῶν κατασκευασθέντων ὁμοίως δείξομεν, ratio to D than C (has) to D [Def. 5.7].
ὅτι τὸ μὲν Ν τοῦ Κ ὑπερέχει, τὸ δὲ Ν τοῦ ΖΘ οὐχ ὑπερέχει. So, I say that D also has a greater ratio to C than D
καί ἐστι τὸ μὲν Ν τοῦ Δ πολλαπλάσιον, τὰ δὲ ΖΘ, Κ τῶν (has) to AB.
ΑΒ, Γ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια· τὸ Δ ἄρα πρὸς For, similarly, by the same construction, we can show
τὸ Γ μείζονα λόγον ἔχει ἤπερ τὸ Δ πρὸς τὸ ΑΒ. that N exceeds K, and N does not exceed FH. And
᾿Αλλὰ δὴ τὸ ΑΕ τοῦ ΕΒ μεῖζον ἔστω. τὸ δὴ ἔλαττον N is a multiple of D, and FH, K other random equal
τὸ ΕΒ πολλαπλασιαζόμενον ἔσται ποτὲ τοῦ Δ μεῖζον. πε- multiples of AB, C (respectively). Thus, D has a greater


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πολλαπλασιάσθω, καὶ ἔστω τὸ ΗΘ πολλαπλάσιον μὲν τοῦ ratio to C than D (has) to AB [Def. 5.5].
ΕΒ, μεῖζον δὲ τοῦ Δ· καὶ ὁσαπλάσιόν ἐστι τὸ ΗΘ τοῦ ΕΒ, And so let AE be greater than EB. So, the lesser,
τοσαυταπλάσιον γεγονέτω καὶ τὸ μὲν ΖΗ τοῦ ΑΕ, τὸ δὲ Κ EB, being multiplied, will sometimes be greater than D.
τοῦ Γ. ὁμοίως δὴ δείξομεν, ὅτι τὰ ΖΘ, Κ τῶν ΑΒ, Γ ἰσάκις Let it have been multiplied, and let GH be a multiple of
ἐστὶ πολλαπλάσια· καὶ εἰλήφθω ὁμοίως τὸ Ν πολλαπλάσιον EB (which is) greater than D. And as many times as
μὲν τοῦ Δ, πρώτως δὲ μεῖζον τοῦ ΖΗ· ὥστε πάλιν τὸ ΖΗ GH is (divisible) by EB, so many times let FG also have
τοῦ Μ οὔκ ἐστιν ἔλασσον. μεῖζον δὲ τὸ ΗΘ τοῦ Δ· ὅλον become (divisible) by AE, and K by C. So, similarly
ἄρα τὸ ΖΘ τῶν Δ, Μ, τουτέστι τοῦ Ν, ὑπερέχει. τὸ δὲ Κ (to the above), we can show that FH and K are equal
τοῦ Ν οὐχ ὑπερέχει, ἐπειδήπερ καὶ τὸ ΖΗ μεῖζον ὂν τοῦ multiples of AB and C (respectively). And, similarly (to
ΗΘ, τουτέστι τοῦ Κ, τοῦ Ν οὐχ ὑπερέχει. καὶ ὡσαύτως the above), let the multiple N of D, (which is) the first
κατακολουθοῦντες τοῖς ἐπάνω περαίνομεν τὴν ἀπόδειξιν. (multiple) greater than FG, have been taken. So, FG
Τῶν ἄρα ἀνίσων μεγεθῶν τὸ μεῖζον πρὸς τὸ αὐτὸ is again not less than M. And GH (is) greater than D.
μείζονα λόγον ἔχει ἤπερ τὸ ἔλαττον· καὶ τὸ αὐτὸ πρὸς τὸ Thus, the whole of FH exceeds D and M, that is to say
ἔλαττον μείζονα λόγον ἔχει ἤπερ πρὸς τὸ μεῖζον· ὅπερ ἔδει N. And K does not exceed N, inasmuch as FG, which
jþ tude) has a greater ratio than the lesser to the same (mag-
δεῖξαι. (is) greater than GH—that is to say, K—also does not
exceed N. And, following the above (arguments), we
(can) complete the proof in the same manner.
Thus, for unequal magnitudes, the greater (magni-

nitude). And the latter (magnitude) has a greater ratio to
the lesser (magnitude) than to the greater. (Which is) the
very thing it was required to show.

Proposition 9
.
Τὰ πρὸς τὸ αὐτὸ τὸν αὐτὸν ἔχοντα λὸγον ἴσα ἀλλήλοις (Magnitudes) having the same ratio to the same
ἐστίν· καὶ πρὸς ἃ τὸ αὐτὸ τὸν αὐτὸν ἕχει λόγον, ἐκεῖνα ἴσα (magnitude) are equal to one another. And those (mag-
ἐστίν. nitudes) to which the same (magnitude) has the same
ratio are equal.
Α Β A B

Γ C

᾿Εχέτω γὰρ ἑκάτερον τῶν Α, Β πρὸς τὸ Γ τὸν αὐτὸν For let A and B each have the same ratio to C. I say
λόγον· λέγω, ὅτι ἴσον ἐστὶ τὸ Α τῷ Β. that A is equal to B.
Εἰ γὰρ μή, οὐκ ἂν ἑκάτερον τῶν Α, Β πρὸς τὸ Γ τὸν For if not, A and B would not each have the same
αὐτὸν εἶχε λόγον· ἔχει δέ· ἴσον ἄρα ἐστὶ τὸ Α τῷ Β. ratio to C [Prop. 5.8]. But they do. Thus, A is equal to
᾿Εχέτω δὴ πάλιν τὸ Γ πρὸς ἑκάτερον τῶν Α, Β τὸν αὐτὸν B.
λόγον· λέγω, ὅτι ἴσον ἐστὶ τὸ Α τῷ Β. So, again, let C have the same ratio to each of A and
iþ tudes) to which the same (magnitude) has the same ratio
Εἰ γὰρ μή, οὐκ ἂν τὸ Γ πρὸς ἑκάτερον τῶν Α, Β τὸν B. I say that A is equal to B.
αὐτὸν εἶχε λόγον· ἔχει δέ· ἴσον ἄρα ἐστὶ τὸ Α τῷ Β. For if not, C would not have the same ratio to each of
Τὰ ἄρα πρὸς τὸ αὐτὸ τὸν αὐτὸν ἔχοντα λόγον ἴσα A and B [Prop. 5.8]. But it does. Thus, A is equal to B.
ἀλλήλοις ἐστίν· καὶ πρὸς ἃ τὸ αὐτὸ τὸν αὐτὸν ἔχει λόγον, Thus, (magnitudes) having the same ratio to the same
ἐκεῖνα ἴσα ἐστίν· ὅπερ ἔδει δεῖξαι. (magnitude) are equal to one another. And those (magni-


are equal. (Which is) the very thing it was required to
show.

Proposition 10
.
Τῶν πρὸς τὸ αὐτὸ λόγον ἐχόντων τὸ μείζονα λόγον For (magnitudes) having a ratio to the same (mag-
ἔχον ἐκεῖνο μεῖζόν ἐστιν· πρὸς ὃ δὲ τὸ αὐτὸ μείζονα λόγον nitude), that (magnitude which) has the greater ratio is


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ἔχει, ἐκεῖνο ἔλαττόν ἐστιν. (the) greater. And that (magnitude) to which the latter
(magnitude) has a greater ratio is (the) lesser.
Α Β A B

Γ C
᾿Εχέτω γὰρ τὸ Α πρὸς τὸ Γ μείζονα λόγον ἤπερ τὸ Β For let A have a greater ratio to C than B (has) to C.
πρὸς τὸ Γ· λέγω, ὅτι μεῖζόν ἐστι τὸ Α τοῦ Β. I say that A is greater than B.
Εἰ γὰρ μή, ἤτοι ἴσον ἐστὶ τὸ Α τῷ Β ἢ ἔλασσον. ἴσον For if not, A is surely either equal to or less than B.
μὲν οὖν οὔκ ἐστὶ τὸ Α τῷ Β· ἑκάτερον γὰρ ἂν τῶν Α, Β In fact, A is not equal to B. For (then) A and B would
πρὸς τὸ Γ τὸν αὐτὸν εἶχε λόγον. οὐκ ἔχει δέ· οὐκ ἄρα ἴσον each have the same ratio to C [Prop. 5.7]. But they do
ἐστὶ τὸ Α τῷ Β. οὐδὲ μὴν ἔλασσόν ἐστι τὸ Α τοῦ Β· τὸ Α not. Thus, A is not equal to B. Neither, indeed, is A less
γὰρ ἂν πρὸς τὸ Γ ἐλάσσονα λόγον εἶχεν ἤπερ τὸ Β πρὸς τὸ than B. For (then) A would have a lesser ratio to C than
Γ. οὐκ ἔχει δέ· οὐκ ἄρα ἔλασσόν ἐστι τὸ Α τοῦ Β. ἐδείχθη B (has) to C [Prop. 5.8]. But it does not. Thus, A is not
δὲ οὐδὲ ἴσον· μεῖζον ἄρα ἐστὶ τὸ Α τοῦ Β. less than B. And it was shown not (to be) equal either.
᾿Εχέτω δὴ πάλιν τὸ Γ πρὸς τὸ Β μείζονα λόγον ἤπερ τὸ Thus, A is greater than B.
Γ πρὸς τὸ Α· λέγω, ὅτι ἔλασσόν ἐστι τὸ Β τοῦ Α. So, again, let C have a greater ratio to B than C (has)
Εἰ γὰρ μή, ἤτοι ἴσον ἐστὶν ἢ μεῖζον. ἴσον μὲν οὖν οὔκ to A. I say that B is less than A.
ἐστι τὸ Β τῷ Α· τὸ Γ γὰρ ἂν πρὸς ἑκάτερον τῶν Α, Β τὸν For if not, (it is) surely either equal or greater. In fact,
αὐτὸν εἶχε λόγον. οὐκ ἔχει δέ· οὐκ ἄρα ἴσον ἐστὶ τὸ Α B is not equal to A. For (then) C would have the same
τῷ Β. οὐδὲ μὴν μεῖζόν ἐστι τὸ Β τοῦ Α· τὸ Γ γὰρ ἂν πρὸς ratio to each of A and B [Prop. 5.7]. But it does not.
τὸ Β ἐλάσσονα λόγον εἶχεν ἤπερ πρὸς τὸ Α. οὐκ ἔχει δέ· Thus, A is not equal to B. Neither, indeed, is B greater
οὐκ ἄρα μεῖζόν ἐστι τὸ Β τοῦ Α. ἐδείχθη δέ, ὅτι οὐδὲ ἴσον· than A. For (then) C would have a lesser ratio to B than
iaþ (magnitude), that (magnitude which) has the greater
ἔλαττον ἄρα ἐστὶ τὸ Β τοῦ Α. (it has) to A [Prop. 5.8]. But it does not. Thus, B is not
Τῶν ἄρα πρὸς τὸ αὐτὸ λόγον ἐχόντων τὸ μείζονα λόγον greater than A. And it was shown that (it is) not equal
ἔχον μεῖζόν ἐστιν· καὶ πρὸς ὃ τὸ αὐτὸ μείζονα λόγον ἔχει, (to A) either. Thus, B is less than A.
ἐκεῖνο ἔλαττόν ἐστιν· ὅπερ ἔδει δεῖξαι. Thus, for (magnitudes) having a ratio to the same
ratio is (the) greater. And that (magnitude) to which
the latter (magnitude) has a greater ratio is (the) lesser.
(Which is) the very thing it was required to show.
.
Proposition 11
†
Οἱ τῷ αὐτῷ λόγῳ οἱ αὐτοὶ καὶ ἀλλήλοις εἰσὶν οἱ αὐτοί. (Ratios which are) the same with the same ratio are
also the same with one another.
Α Γ Ε A C E
Β ∆ Ζ B D F
Η Θ Κ G H K
Λ Μ Ν L M N
῎Εστωσαν γὰρ ὡς μὲν τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς For let it be that as A (is) to B, so C (is) to D, and as
τὸ Δ, ὡς δὲ τὸ Γ πρὸς τὸ Δ, οὕτως τὸ Ε πρὸς τὸ Ζ· λέγω, C (is) to D, so E (is) to F. I say that as A is to B, so E
ὅτι ἐστὶν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Ε πρὸς τὸ Ζ. (is) to F.
Εἰλήφθω γὰρ τῶν Α, Γ, Ε ἰσάκις πολλαπλάσια τὰ Η, Θ, For let the equal multiples G, H, K have been taken
Κ, τῶν δὲ Β, Δ, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ of A, C, E (respectively), and the other random equal
Λ, Μ, Ν. multiples L, M, N of B, D, F (respectively).
Καὶ ἐπεί ἐστιν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ And since as A is to B, so C (is) to D, and the equal
Δ, καὶ εἴληπται τῶν μὲν Α, Γ ἰσάκις πολλαπλάσια τὰ Η, Θ, multiples G and H have been taken of A and C (respec-
τῶν δὲ Β, Δ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Λ, Μ, tively), and the other random equal multiples L and M
εἰ ἄρα ὑπερέχει τὸ Η τοῦ Λ, ὑπερέχει καὶ τὸ Θ τοῦ Μ, καὶ εἰ of B and D (respectively), thus if G exceeds L then H
ἴσον ἐστίν, ἴσον, καὶ εἰ ἐλλείπει, ἐλλείπει. πάλιν, ἐπεί ἐστιν also exceeds M, and if (G is) equal (to L then H is also)


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ὡς τὸ Γ πρὸς τὸ Δ, οὕτως τὸ Ε πρὸς τὸ Ζ, καὶ εἴληπται equal (to M), and if (G is) less (than L then H is also)
τῶν Γ, Ε ἰσάκις πολλαπλάσια τὰ Θ, Κ, τῶν δὲ Δ, Ζ ἄλλα, less (than M) [Def. 5.5]. Again, since as C is to D, so
ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Μ, Ν, εἰ ἄρα ὑπερέχει τὸ E (is) to F, and the equal multiples H and K have been
Θ τοῦ Μ, ὑπερέχει καὶ τὸ Κ τοῦ Ν, καὶ εἰ ἴσον, ἴσον, καὶ εἰ taken of C and E (respectively), and the other random
ἔλλατον, ἔλαττον. ἀλλὰ εἰ ὑπερεῖχε τὸ Θ τοῦ Μ, ὑπερεῖχε equal multiples M and N of D and F (respectively), thus
καὶ τὸ Η τοῦ Λ, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον· if H exceeds M then K also exceeds N, and if (H is)
ὥστε καὶ εἰ ὑπερέχει τὸ Η τοῦ Λ, ὑπερέχει καὶ τὸ Κ τοῦ equal (to M then K is also) equal (to N), and if (H is)
Ν, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον. καί ἐστι τὰ less (than M then K is also) less (than N) [Def. 5.5]. But
μὲν Η, Κ τῶν Α, Ε ἰσάκις πολλαπλάσια, τὰ δὲ Λ, Ν τῶν Β, (we saw that) if H was exceeding M then G was also ex-
Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια· ἔστιν ἄρα ὡς τὸ Α ceeding L, and if (H was) equal (to M then G was also)
πρὸς τὸ Β, οὕτως τὸ Ε πρὸς τὸ Ζ. equal (to L), and if (H was) less (than M then G was
Οἱ ἄρα τῷ αὐτῷ λόγῳ οἱ αὐτοὶ καὶ ἀλλήλοις εἰσὶν οἱ also) less (than L). And, hence, if G exceeds L then K
αὐτοί· ὅπερ ἔδει δεῖξαι. also exceeds N, and if (G is) equal (to L then K is also)
equal (to N), and if (G is) less (than L then K is also)
less (than N). And G and K are equal multiples of A
ibþ so E (is) to F [Def. 5.5].
and E (respectively), and L and N other random equal
multiples of B and F (respectively). Thus, as A is to B,
Thus, (ratios which are) the same with the same ratio
are also the same with one another. (Which is) the very
thing it was required to show.
† In modern notation, this proposition reads that if α : β :: γ : δ and γ : δ :: ǫ : ζ then α : β :: ǫ : ζ.

.
Proposition 12
†
᾿Εὰν ᾖ ὁποσαοῦν μεγέθη ἀνάλογον, ἔσται ὡς ἓν τῶν If there are any number of magnitudes whatsoever
ἡγουμένων πρὸς ἓν τῶν ἑπομένων, οὕτως ἅπαντα τὰ (which are) proportional then as one of the leading (mag-
ἡγούμενα πρὸς ἅπαντα τὰ ἑπόμενα. nitudes is) to one of the following, so will all of the lead-
ing (magnitudes) be to all of the following.
Α Γ Ε A C E

Β ∆ Ζ B D F

Η Λ G L

Θ Μ H M
Κ Ν K N
῎Εστωσαν ὁποσαοῦν μεγέθη ἀνάλογον τὰ Α, Β, Γ, Δ, Let there be any number of magnitudes whatsoever,
Ε, Ζ, ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ, καὶ τὸ Ε A, B, C, D, E, F, (which are) proportional, (so that) as
πρὸς το Ζ· λέγω, ὅτι ἐστὶν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὰ A (is) to B, so C (is) to D, and E to F. I say that as A is
Α, Γ, Ε πρὸς τὰ Β, Δ, Ζ. to B, so A, C, E (are) to B, D, F.
Εἰλήφθω γὰρ τῶν μὲν Α, Γ, Ε ἰσάκις πολλαπλάσια τὰ Η, For let the equal multiples G, H, K have been taken
Θ, Κ, τῶν δὲ Β, Δ, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια of A, C, E (respectively), and the other random equal
τὰ Λ, Μ, Ν. multiples L, M, N of B, D, F (respectively).
Καὶ ἐπεί ἐστιν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ And since as A is to B, so C (is) to D, and E to F, and
Δ, καὶ τὸ Ε πρὸς τὸ Ζ, καὶ εἴληπται τῶν μὲν Α, Γ, Ε ἰσάκις the equal multiples G, H, K have been taken of A, C, E
πολλαπλάσια τὰ Η, Θ, Κ τῶν δὲ Β, Δ, Ζ ἄλλα, ἃ ἔτυχεν, (respectively), and the other random equal multiples L,
ἰσάκις πολλαπλάσια τὰ Λ, Μ, Ν, εἰ ἄρα ὑπερέχει τὸ Η τοῦ Λ, M, N of B, D, F (respectively), thus if G exceeds L then
ὑπερέχει καὶ τὸ Θ τοῦ Μ, καὶ τὸ Κ τοῦ Ν, καὶ εἰ ἴσον, ἴσον, H also exceeds M, and K (exceeds) N, and if (G is)
καὶ εἰ ἔλαττον, ἔλαττον. ὥστε καὶ εἰ ὑπερέχει τὸ Η τοῦ Λ, equal (to L then H is also) equal (to M, and K to N),


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ὑπερέχει καὶ τὰ Η, Θ, Κ τῶν Λ, Μ, Ν, καὶ εἰ ἴσον, ἴσα, καὶ and if (G is) less (than L then H is also) less (than M,
εἰ ἔλαττον, ἔλαττονα. καί ἐστι τὸ μὲν Η καὶ τὰ Η, Θ, Κ and K than N) [Def. 5.5]. And, hence, if G exceeds L
τοῦ Α καὶ τῶν Α, Γ, Ε ἰσάκις πολλαπλάσια, ἐπειδήπερ ἐὰν then G, H, K also exceed L, M, N, and if (G is) equal
ᾖ ὁποσαοῦν μεγέθη ὁποσωνοῦν μεγεθῶν ἴσων τὸ πλῆθος (to L then G, H, K are also) equal (to L, M, N) and
ἕκαστον ἑκάστου ἰσάκις πολλαπλάσιον, ὁσαπλάσιόν ἐστιν if (G is) less (than L then G, H, K are also) less (than
ἓν τῶν μεγεθῶν ἑνός, τοσαυταπλάσια ἔσται καὶ τὰ πάντα L, M, N). And G and G, H, K are equal multiples of
τῶν πάντων. διὰ τὰ αὐτὰ δὴ καὶ τὸ Λ καὶ τὰ Λ, Μ, Ν τοῦ A and A, C, E (respectively), inasmuch as if there are
Β καὶ τῶν Β, Δ, Ζ ἰσάκις ἐστὶ πολλαπλάσια· ἔστιν ἄρα ὡς any number of magnitudes whatsoever (which are) equal
τὸ Α πρὸς τὸ Β, οὕτως τὰ Α, Γ, Ε πρὸς τὰ Β, Δ, Ζ. multiples, respectively, of some (other) magnitudes, of
᾿Εὰν ἄρα ᾖ ὁποσαοῦν μεγέθη ἀνάλογον, ἔσται ὡς ἓν equal number (to them), then as many times as one of the
τῶν ἡγουμένων πρὸς ἓν τῶν ἑπομένων, οὕτως ἅπαντα τὰ (first) magnitudes is (divisible) by one (of the second),
ἡγούμενα πρὸς ἅπαντα τὰ ἑπόμενα· ὅπερ ἔδει δεῖξαι. so many times will all (of the first magnitudes) also (be
divisible) by all (of the second) [Prop. 5.1]. So, for the
same (reasons), L and L, M, N are also equal multiples
of B and B, D, F (respectively). Thus, as A is to B, so
igþ ever (which are) proportional then as one of the leading
A, C, E (are) to B, D, F (respectively).
Thus, if there are any number of magnitudes whatso-
(magnitudes is) to one of the following, so will all of the
leading (magnitudes) be to all of the following. (Which
is) the very thing it was required to show.
′
′
′
′
† In modern notation, this proposition reads that if α : α :: β : β :: γ : γ etc. then α : α :: (α + β + γ + · · · ) : (α + β + γ + · · · ).
′
′
′
Proposition 13
.
†
᾿Εὰν πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἒχῃ λόγον καὶ If a first (magnitude) has the same ratio to a second
τρίτον πρὸς τέταρτον, τρίτον δὲ πρὸς τέταρτον μείζονα that a third (has) to a fourth, and the third (magnitude)
λόγον ἔχῃ ἢ πέμπτον πρὸς ἕκτον, καὶ πρῶτον πρὸς δεύτερον has a greater ratio to the fourth than a fifth (has) to a
μείζονα λόγον ἕξει ἢ πέμπτον πρὸς ἕκτον. sixth, then the first (magnitude) will also have a greater
ratio to the second than the fifth (has) to the sixth.
Α Γ Ε A C E
Β ∆ Ζ B D F
Μ Η Θ M G H
Ν Κ Λ N K L
Πρῶτον γὰρ τὸ Α πρὸς δεύτερον τὸ Β τὸν αὐτὸν ἐχέτω For let a first (magnitude) A have the same ratio to a
λόγον καὶ τρίτον τὸ Γ πρὸς τέταρτον τὸ Δ, τρίτον δὲ τὸ Γ second B that a third C (has) to a fourth D, and let the
πρὸς τέταρτον τὸ Δ μείζονα λόγον ἐχέτω ἢ πέμπτον τὸ Ε third (magnitude) C have a greater ratio to the fourth
πρὸς ἕκτον τὸ Ζ. λέγω, ὅτι καὶ πρῶτον τὸ Α πρὸς δεύτερον D than a fifth E (has) to a sixth F. I say that the first
τὸ Β μείζονα λόγον ἕξει ἤπερ πέμπτον τὸ Ε πρὸς ἕκτον τὸ (magnitude) A will also have a greater ratio to the second
Ζ. B than the fifth E (has) to the sixth F.
᾿Επεὶ γὰρ ἔστι τινὰ τῶν μὲν Γ, Ε ἰσάκις πολλαπλάσια, For since there are some equal multiples of C and
τῶν δὲ Δ, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια, καὶ τὸ μὲν E, and other random equal multiples of D and F, (for
τοῦ Γ πολλαπλάσιον τοῦ τοῦ Δ πολλαπλασίου ὑπερέχει, which) the multiple of C exceeds the (multiple) of D,
τὸ δὲ τοῦ Ε πολλαπλάσιον τοῦ τοῦ Ζ πολλαπλασίου οὐχ and the multiple of E does not exceed the multiple of F
ὑπερέχει, εἰλήφθω, καὶ ἔστω τῶν μὲν Γ, Ε ἰσάκις πολ- [Def. 5.7], let them have been taken. And let G and H be
λαπλάσια τὰ Η, Θ, τῶν δὲ Δ, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις equal multiples of C and E (respectively), and K and L
πολλαπλάσια τὰ Κ, Λ, ὥστε τὸ μὲν Η τοῦ Κ ὑπερέχειν, τὸ other random equal multiples of D and F (respectively),
δὲ Θ τοῦ Λ μὴ ὑπερέχειν· καὶ ὁσαπλάσιον μέν ἐστι τὸ Η such that G exceeds K, but H does not exceed L. And as
τοῦ Γ, τοσαυταπλάσιον ἔστω καὶ τὸ Μ τοῦ Α, ὁσαπλάσιον many times as G is (divisible) by C, so many times let M
δὲ τὸ Κ τοῦ Δ, τοσαυταπλάσιον ἔστω καὶ τὸ Ν τοῦ Β. be (divisible) by A. And as many times as K (is divisible)


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Καὶ ἐπεί ἐστιν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς by D, so many times let N be (divisible) by B.
τὸ Δ, καὶ εἴληπται τῶν μὲν Α, Γ ἰσάκις πολλαπλάσια τὰ And since as A is to B, so C (is) to D, and the equal
Μ, Η, τῶν δὲ Β, Δ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ multiples M and G have been taken of A and C (respec-
Ν, Κ, εἰ ἄρα ὑπερέχει τὸ Μ τοῦ Ν, ὑπερέχει καὶ τὸ Η τοῦ tively), and the other random equal multiples N and K
Κ, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλλατον. ὑπερέχει δὲ of B and D (respectively), thus if M exceeds N then G
τὸ Η τοῦ Κ· ὑπερέχει ἄρα καὶ τὸ Μ τοῦ Ν. τὸ δὲ Θ τοῦ exceeds K, and if (M is) equal (to N then G is also)
Λ οὐχ ὑπερέχει· καί ἐστι τὰ μὲν Μ, Θ τῶν Α, Ε ἰσάκις equal (to K), and if (M is) less (than N then G is also)
πολλαπλάσια, τὰ δὲ Ν, Λ τῶν Β, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις less (than K) [Def. 5.5]. And G exceeds K. Thus, M
πολλαπλάσια· τὸ ἄρα Α πρὸς τὸ Β μείζονα λόγον ἔχει ἤπερ also exceeds N. And H does not exceeds L. And M and
τὸ Ε πρὸς τὸ Ζ. H are equal multiples of A and E (respectively), and N
᾿Εὰν ἄρα πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἒχῃ λόγον and L other random equal multiples of B and F (respec-
καὶ τρίτον πρὸς τέταρτον, τρίτον δὲ πρὸς τέταρτον μείζονα tively). Thus, A has a greater ratio to B than E (has) to
λόγον ἔχῃ ἢ πέμπτον πρὸς ἕκτον, καὶ πρῶτον πρὸς δεύτερον F [Def. 5.7].
idþ to a sixth, then the first (magnitude) will also have a
μείζονα λόγον ἕξει ἢ πέμπτον πρὸς ἕκτον· ὅπερ ἔδει δεῖξαι. second that a third (has) to a fourth, and a third (magni-
Thus, if a first (magnitude) has the same ratio to a
tude) has a greater ratio to a fourth than a fifth (has)

greater ratio to the second than the fifth (has) to the
sixth. (Which is) the very thing it was required to show.

† In modern notation, this proposition reads that if α : β :: γ : δ and γ : δ > ǫ : ζ then α : β > ǫ : ζ. †

.
Proposition 14
᾿Εὰν πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἔχῃ λόγον καὶ If a first (magnitude) has the same ratio to a second
τρίτον πρὸς τέταρτον, τὸ δὲ πρῶτον τοῦ τρίτου μεῖζον ᾖ, that a third (has) to a fourth, and the first (magnitude)
καὶ τὸ δεύτερον τοῦ τετάρτου μεῖζον ἔσται, κἂν ἴσον, ἴσον, is greater than the third, then the second will also be
κἂν ἔλαττον, ἔλαττον. greater than the fourth. And if (the first magnitude is)
equal (to the third then the second will also be) equal (to
the fourth). And if (the first magnitude is) less (than the
third then the second will also be) less (than the fourth).
Α Γ A C

Β ∆ B D
Πρῶτον γὰρ τὸ Α πρὸς δεύτερον τὸ Β αὐτὸν ἐχέτω For let a first (magnitude) A have the same ratio to a
λόγον καὶ τρίτον τὸ Γ πρὸς τέταρτον τὸ Δ, μεῖζον δὲ ἔστω second B that a third C (has) to a fourth D. And let A be
τὸ Α τοῦ Γ· λέγω, ὅτι καὶ τὸ Β τοῦ Δ μεῖζόν ἐστιν. greater than C. I say that B is also greater than D.
᾿Επεὶ γὰρ τὸ Α τοῦ Γ μεῖζόν ἐστιν, ἄλλο δέ, ὃ ἔτυχεν, For since A is greater than C, and B (is) another ran-
[μέγεθος] τὸ Β, τὸ Α ἄρα πρὸς τὸ Β μείζονα λόγον ἔχει dom [magnitude], A thus has a greater ratio to B than C
ἤπερ τὸ Γ πρὸς τὸ Β. ὡς δὲ τὸ Α πρὸς τὸ Β, οὕτως τὸ (has) to B [Prop. 5.8]. And as A (is) to B, so C (is) to
Γ πρὸς τὸ Δ· καὶ τὸ Γ ἄρα πρὸς τὸ Δ μείζονα λόγον ἔχει D. Thus, C also has a greater ratio to D than C (has) to
ἤπερ τὸ Γ πρὸς τὸ Β. πρὸς ὃ δὲ τὸ αὐτὸ μείζονα λόγον B. And that (magnitude) to which the same (magnitude)
ἔχει, ἐκεῖνο ἔλασσόν ἐστιν· ἔλασσον ἄρα τὸ Δ τοῦ Β· ὥστε has a greater ratio is the lesser [Prop. 5.10]. Thus, D (is)
μεῖζόν ἐστι τὸ Β τοῦ Δ. less than B. Hence, B is greater than D.
῾Ομοίως δὴ δεῖξομεν, ὅτι κἂν ἴσον ᾖ τὸ Α τῷ Γ, ἴσον So, similarly, we can show that even if A is equal to C
ἔσται καὶ τὸ Β τῷ Δ, κἄν ἔλασσον ᾖ τὸ Α τοῦ Γ, ἔλασσον then B will also be equal to D, and even if A is less than
ἔσται καὶ τὸ Β τοῦ Δ. C then B will also be less than D.
᾿Εὰν ἄρα πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἔχῃ λόγον Thus, if a first (magnitude) has the same ratio to a
καὶ τρίτον πρὸς τέταρτον, τὸ δὲ πρῶτον τοῦ τρίτου μεῖζον ᾖ, second that a third (has) to a fourth, and the first (mag-
καὶ τὸ δεύτερον τοῦ τετάρτου μεῖζον ἔσται, κἂν ἴσον, ἴσον, nitude) is greater than the third, then the second will also
κἂν ἔλαττον, ἔλαττον· ὅπερ ἔδει δεῖξαι. be greater than the fourth. And if (the first magnitude is)


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ieþ equal (to the third then the second will also be) equal (to



the fourth). And if (the first magnitude is) less (than the
third then the second will also be) less (than the fourth).
(Which is) the very thing it was required to show.

† In modern notation, this proposition reads that if α : β :: γ : δ then α T γ as β T δ.

Proposition 15
.
†
Τὰ μέρη τοῖς ὡσαύτως πολλαπλασίοις τὸν αὐτὸν ἔχει Parts have the same ratio as similar multiples, taken
λόγον ληφθέντα κατάλληλα. in corresponding order.
Α Η Θ Β A G H B
Γ C

∆ Κ Λ Ε D K L E
Ζ F
῎Εστω γὰρ ἰσάκις πολλαπλάσιον τὸ ΑΒ τοῦ Γ καὶ το ΔΕ For let AB and DE be equal multiples of C and F
τοῦ Ζ· λέγω, ὅτι ἐστὶν ὡς τὸ Γ πρὸς τὸ Ζ, οὕτως τὸ ΑΒ (respectively). I say that as C is to F, so AB (is) to DE.
πρὸς τὸ ΔΕ. For since AB and DE are equal multiples of C and
᾿Επεὶ γὰρ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΑΒ τοῦ Γ καὶ F (respectively), thus as many magnitudes as there are
τὸ ΔΕ τοῦ Ζ, ὅσα ἄρα ἐστὶν ἐν τῷ ΑΒ μεγέθη ἴσα τῷ in AB equal to C, so many (are there) also in DE equal
Γ, τοσαῦτα καὶ ἐν τῷ ΔΕ ἴσα τῷ Ζ. διῃρήσθω τὸ μὲν ΑΒ to F. Let AB have been divided into (magnitudes) AG,
εἰς τὰ τῷ Γ ἴσα τὰ ΑΗ, ΗΘ, ΘΒ, τὸ δὲ ΔΕ εἰς τὰ τῷ Ζ GH, HB, equal to C, and DE into (magnitudes) DK,
ἴσα τὰ ΔΚ, ΚΛ, ΛΕ· ἔσται δὴ ἴσον τὸ πλῆθος τῶν ΑΗ, KL, LE, equal to F. So, the number of (magnitudes)
ΗΘ, ΘΒ τῷ πλήθει τῶν ΔΚ, ΚΛ, ΛΕ. καὶ ἐπεὶ ἴσα ἐστὶ τὰ AG, GH, HB will equal the number of (magnitudes)
ΑΗ, ΗΘ, ΘΒ ἀλλήλοις, ἔστι δὲ καὶ τὰ ΔΚ, ΚΛ, ΛΕ ἴσα DK, KL, LE. And since AG, GH, HB are equal to one
ἀλλήλοις, ἔστιν ἄρα ὡς τὸ ΑΗ πρὸς τὸ ΔΚ, οὕτως τὸ ΗΘ another, and DK, KL, LE are also equal to one another,
πρὸς τὸ ΚΛ, καὶ τὸ ΘΒ πρὸς τὸ ΛΕ. ἔσται ἄρα καὶ ὡς ἓν thus as AG is to DK, so GH (is) to KL, and HB to LE
τῶν ἡγουμένων πρὸς ἓν τῶν ἑπομένων, οὕτως ἅπαντα τὰ [Prop. 5.7]. And, thus (for proportional magnitudes), as
ἡγουμένα πρὸς ἅπαντα τὰ ἑπόμενα· ἔστιν ἄρα ὡς τὸ ΑΗ one of the leading (magnitudes) will be to one of the fol-
πρὸς τὸ ΔΚ, οὕτως τὸ ΑΒ πρὸς τὸ ΔΕ. ἴσον δὲ τὸ μὲν ΑΗ lowing, so all of the leading (magnitudes will be) to all of
iþ taken in corresponding order. (Which is) the very thing
τῷ Γ, τὸ δὲ ΔΚ τῷ Ζ· ἔστιν ἄρα ὡς τὸ Γ πρὸς τὸ Ζ οὕτως the following [Prop. 5.12]. Thus, as AG is to DK, so AB
τὸ ΑΒ πρὸς τὸ ΔΕ. (is) to DE. And AG is equal to C, and DK to F. Thus,
Τὰ ἄρα μέρη τοῖς ὡσαύτως πολλαπλασίοις τὸν αὐτὸν as C is to F, so AB (is) to DE.
ἔχει λόγον ληφθέντα κατάλληλα· ὅπερ ἔδει δεῖξαι. Thus, parts have the same ratio as similar multiples,

it was required to show.

† In modern notation, this proposition reads that α : β :: m α : m β. †

.
Proposition 16
᾿Εὰν τέσσαρα μεγέθη ἀνάλογον ᾖ, καὶ ἐναλλὰξ ἀνάλογον If four magnitudes are proportional then they will also
ἔσται. be proportional alternately.
῎Εστω τέσσαρα μεγέθη ἀνάλογον τὰ Α, Β, Γ, Δ, ὡς τὸ Let A, B, C and D be four proportional magnitudes,
Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ· λέγω, ὅτι καὶ ἐναλλὰξ (such that) as A (is) to B, so C (is) to D. I say that they
[ἀνάλογον] ἔσται, ὡς τὸ Α πρὸς τὸ Γ, οὕτως τὸ Β πρὸς τὸ will also be [proportional] alternately, (so that) as A (is)
Δ. to C, so B (is) to D.
Εἰλήφθω γὰρ τῶν μὲν Α, Β ἰσάκις πολλαπλάσια τὰ Ε, For let the equal multiples E and F have been taken
Ζ, τῶν δὲ Γ, Δ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Η, of A and B (respectively), and the other random equal
Θ. multiples G and H of C and D (respectively).


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C
Α Γ A ELEMENTS BOOK 5
Β ∆ B D
Ε Η E G

Ζ Θ F H
Καὶ ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ Ε τοῦ Α καὶ τὸ Ζ And since E and F are equal multiples of A and B
τοῦ Β, τὰ δὲ μέρη τοῖς ὡσαύτως πολλαπλασίοις τὸν αὐτὸν (respectively), and parts have the same ratio as similar
ἔχει λόγον, ἔστιν ἄρα ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Ε πρὸς multiples [Prop. 5.15], thus as A is to B, so E (is) to F.
τὸ Ζ. ὡς δὲ τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ· καὶ ὡς But as A (is) to B, so C (is) to D. And, thus, as C (is)
ἄρα τὸ Γ πρὸς τὸ Δ, οὕτως τὸ Ε πρὸς τὸ Ζ. πάλιν, ἐπεὶ τὰ to D, so E (is) to F [Prop. 5.11]. Again, since G and H
Η, Θ τῶν Γ, Δ ἰσάκις ἐστὶ πολλαπλάσια, ἔστιν ἄρα ὡς τὸ Γ are equal multiples of C and D (respectively), thus as C
πρὸς τὸ Δ, οὕτως τὸ Η πρὸς τὸ Θ. ὡς δὲ τὸ Γ πρὸς τὸ Δ, is to D, so G (is) to H [Prop. 5.15]. But as C (is) to D,
[οὕτως] τὸ Ε πρὸς τὸ Ζ· καὶ ὡς ἄρα τὸ Ε πρὸς τὸ Ζ, οὕτως [so] E (is) to F. And, thus, as E (is) to F, so G (is) to
τὸ Η πρὸς τὸ Θ. ἐὰν δὲ τέσσαρα μεγέθη ἀνάλογον ᾖ, τὸ δὲ H [Prop. 5.11]. And if four magnitudes are proportional,
πρῶτον τοῦ τρίτου μεῖζον ᾖ, καὶ τὸ δεύτερον τοῦ τετάρτου and the first is greater than the third then the second will
μεῖζον ἔσται, κἂν ἴσον, ἴσον, κἄν ἔλαττον, ἔλαττον. εἰ ἄρα also be greater than the fourth, and if (the first is) equal
ὑπερέχει τὸ Ε τοῦ Η, ὑπερέχει καὶ τὸ Ζ τοῦ Θ, καὶ εἰ ἴσον, (to the third then the second will also be) equal (to the
ἴσον, καὶ εἰ ἔλαττον, ἔλαττον. καί ἐστι τὰ μὲν Ε, Ζ τῶν fourth), and if (the first is) less (than the third then the
Α, Β ἰσάκις πολλαπλάσια, τὰ δὲ Η, Θ τῶν Γ, Δ ἄλλα, ἃ second will also be) less (than the fourth) [Prop. 5.14].
ἔτυχεν, ἰσάκις πολλαπλάσια· ἔστιν ἄρα ὡς τὸ Α πρὸς τὸ Γ, Thus, if E exceeds G then F also exceeds H, and if (E is)
οὕτως τὸ Β πρὸς τὸ Δ. equal (to G then F is also) equal (to H), and if (E is) less
᾿Εὰν ἄρα τέσσαρα μεγέθη ἀνάλογον ᾖ, καὶ ἐναλλὰξ (than G then F is also) less (than H). And E and F are
izþ will also be proportional alternately. (Which is) the very
ἀνάλογον ἔσται· ὅπερ ἔδει δεῖξαι. equal multiples of A and B (respectively), and G and H
other random equal multiples of C and D (respectively).
Thus, as A is to C, so B (is) to D [Def. 5.5].
Thus, if four magnitudes are proportional then they

thing it was required to show.
† In modern notation, this proposition reads that if α : β :: γ : δ then α : γ :: β : δ.

Proposition 17
†
.
᾿Εὰν συγκείμενα μεγέθη ἀνάλογον ᾖ, καὶ διαιρεθέντα If composed magnitudes are proportional then they
ἀνάλογον ἔσται. will also be proportional (when) separarted.
Α Ε Β Γ Ζ ∆ A E B C F D

Η Θ Κ Ξ G H K O

Λ Μ Ν Π L M N P

῎Εστω συγκείμενα μεγέθη ἀνάλογον τὰ ΑΒ, ΒΕ, ΓΔ, Let AB, BE, CD, and DF be composed magnitudes
ΔΖ, ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ πρὸς τὸ ΔΖ· (which are) proportional, (so that) as AB (is) to BE, so
λέγω, ὅτι καὶ διαιρεθέντα ἀνάλογον ἔσται, ὡς τὸ ΑΕ πρὸς CD (is) to DF. I say that they will also be proportional
τὸ ΕΒ, οὕτως τὸ ΓΖ πρὸς τὸ ΔΖ. (when) separated, (so that) as AE (is) to EB, so CF (is)
Εἰλήφθω γὰρ τῶν μὲν ΑΕ, ΕΒ, ΓΖ, ΖΔ ἰσάκις πολ- to DF.
λαπλάσια τὰ ΗΘ, ΘΚ, ΛΜ, ΜΝ, τῶν δὲ ΕΒ, ΖΔ ἄλλα, ἃ For let the equal multiples GH, HK, LM, and MN
ἔτυχεν, ἰσάκις πολλαπλάσια τὰ ΚΞ, ΝΠ. have been taken of AE, EB, CF, and FD (respectively),
Καὶ ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΗΘ τοῦ ΑΕ καὶ and the other random equal multiples KO and NP of
τὸ ΘΚ τοῦ ΕΒ, ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΗΘ τοῦ EB and FD (respectively).


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ΑΕ καὶ τὸ ΗΚ τοῦ ΑΒ. ἰσάκις δέ ἐστι πολλαπλάσιον τὸ ΗΘ And since GH and HK are equal multiples of AE and
τοῦ ΑΕ καὶ τὸ ΛΜ τοῦ ΓΖ· ἰσάκις ἄρα ἐστὶ πολλαπλάσιον EB (respectively), GH and GK are thus equal multiples
τὸ ΗΚ τοῦ ΑΒ καὶ τὸ ΛΜ τοῦ ΓΖ. πάλιν, ἐπεὶ ἰσάκις ἐστὶ of AE and AB (respectively) [Prop. 5.1]. But GH and
πολλαπλάσιον τὸ ΛΜ τοῦ ΓΖ καὶ τὸ ΜΝ τοῦ ΖΔ, ἰσάκις ἄρα LM are equal multiples of AE and CF (respectively).
ἐστὶ πολλαπλάσιον τὸ ΛΜ τοῦ ΓΖ καὶ τὸ ΛΝ τοῦ ΓΔ. ἰσάκις Thus, GK and LM are equal multiples of AB and CF
δὲ ἦν πολλαπλάσιον τὸ ΛΜ τοῦ ΓΖ καὶ τὸ ΗΚ τοῦ ΑΒ· (respectively). Again, since LM and MN are equal mul-
ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΗΚ τοῦ ΑΒ καὶ τὸ ΛΝ τοῦ tiples of CF and FD (respectively), LM and LN are thus
ΓΔ. τὰ ΗΚ, ΛΝ ἄρα τῶν ΑΒ, ΓΔ ἰσάκις ἐστὶ πολλαπλάσια. equal multiples of CF and CD (respectively) [Prop. 5.1].
πάλιν, ἐπεὶ ἰσάκις ἐστὶ πολλαπλασίον τὸ ΘΚ τοῦ ΕΒ καὶ τὸ And LM and GK were equal multiples of CF and AB
ΜΝ τοῦ ΖΔ, ἔστι δὲ καὶ τὸ ΚΞ τοῦ ΕΒ ἰσάκις πολλαπλάσιον (respectively). Thus, GK and LN are equal multiples
καὶ τὸ ΝΠ τοῦ ΖΔ, καὶ συντεθὲν τὸ ΘΞ τοῦ ΕΒ ἰσάκις ἐστὶ of AB and CD (respectively). Thus, GK, LN are equal
πολλαπλάσιον καὶ τὸ ΜΠ τοῦ ΖΔ. καὶ ἐπεί ἐστιν ὡς τὸ ΑΒ multiples of AB, CD. Again, since HK and MN are
πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ πρὸς τὸ ΔΖ, καὶ εἴληπται τῶν equal multiples of EB and FD (respectively), and KO
μὲν ΑΒ, ΓΔ ἰσάκις πολλαπλάσια τὰ ΗΚ, ΛΝ, τῶν δὲ ΕΒ, and NP are also equal multiples of EB and FD (respec-
ΖΔ ἰσάκις πολλαπλάσια τὰ ΘΞ, ΜΠ, εἰ ἄρα ὑπερέχει τὸ tively), then, added together, HO and MP are also equal
ΗΚ τοῦ ΘΞ, ὑπερέχει καὶ τὸ ΛΝ τοῦ ΜΠ, καὶ εἰ ἴσον, ἴσον, multiples of EB and FD (respectively) [Prop. 5.2]. And
καὶ εἰ ἔλαττον, ἔλαττον. ὑπερεχέτω δὴ τὸ ΗΚ τοῦ ΘΞ, since as AB (is) to BE, so CD (is) to DF, and the equal
καὶ κοινοῦ ἀφαιρεθέντος τοῦ ΘΚ ὑπερέχει ἄρα καὶ τὸ ΗΘ multiples GK, LN have been taken of AB, CD, and the
τοῦ ΚΞ. ἀλλα εἰ ὑπερεῖχε τὸ ΗΚ τοῦ ΘΞ ὑπερεῖχε καὶ τὸ equal multiples HO, MP of EB, FD, thus if GK exceeds
ΛΝ τοῦ ΜΠ· ὑπερέχει ἄρα καὶ τὸ ΛΝ τοῦ ΜΠ, καὶ κοινοῦ HO then LN also exceeds MP, and if (GK is) equal (to
ἀφαιρεθέντος τοῦ ΜΝ ὑπερέχει καὶ τὸ ΛΜ τοῦ ΝΠ· ὥστε HO then LN is also) equal (to MP), and if (GK is) less
εἰ ὑπερέχει τὸ ΗΘ τοῦ ΚΞ, ὑπερέχει καὶ τὸ ΛΜ τοῦ ΝΠ. (than HO then LN is also) less (than MP) [Def. 5.5].
ὁμοίως δὴ δεῖξομεν, ὅτι κἂν ἴσον ᾖ τὸ ΗΘ τῷ ΚΞ, ἴσον So let GK exceed HO, and thus, HK being taken away
ἔσται καὶ τὸ ΛΜ τῷ ΝΠ, κἂν ἔλαττον, ἔλαττον. καί ἐστι τὰ from both, GH exceeds KO. But (we saw that) if GK
μὲν ΗΘ, ΛΜ τῶν ΑΕ, ΓΖ ἰσάκις πολλαπλάσια, τὰ δὲ ΚΞ, was exceeding HO then LN was also exceeding MP.
ΝΠ τῶν ΕΒ, ΖΔ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια· ἔστιν Thus, LN also exceeds MP, and, MN being taken away
ἄρα ὡς τὸ ΑΕ πρὸς τὸ ΕΒ, οὕτως τὸ ΓΖ πρὸς τὸ ΖΔ. from both, LM also exceeds NP. Hence, if GH exceeds
᾿Εὰν ἄρα συγκείμενα μεγέθη ἀνάλογον ᾖ, καὶ διαι- KO then LM also exceeds NP. So, similarly, we can
ρεθέντα ἀνάλογον ἔσται· ὅπερ ἔδει δεῖξαι. show that even if GH is equal to KO then LM will also
be equal to NP, and even if (GH is) less (than KO then
LM will also be) less (than NP). And GH, LM are equal
ihþ to FD [Def. 5.5].
multiples of AE, CF, and KO, NP other random equal
multiples of EB, FD. Thus, as AE is to EB, so CF (is)

Thus, if composed magnitudes are proportional then
they will also be proportional (when) separarted. (Which
is) the very thing it was required to show.

† In modern notation, this proposition reads that if α + β : β :: γ + δ : δ then α : β :: γ : δ.

†
Proposition 18
.
᾿Εὰν διῃρημένα μεγέθη ἀνάλογον ᾖ, καὶ συντεθέντα If separated magnitudes are proportional then they
ἀνάλογον ἔσται. will also be proportional (when) composed.
Α Ε Β A E B


Γ Ζ Η ∆ C F G D

῎Εστω διῃρημένα μεγέθη ἀνάλογον τὰ ΑΕ, ΕΒ, ΓΖ, ΖΔ, Let AE, EB, CF, and FD be separated magnitudes
ὡς τὸ ΑΕ πρὸς τὸ ΕΒ, οὕτως τὸ ΓΖ πρὸς τὸ ΖΔ· λέγω, (which are) proportional, (so that) as AE (is) to EB, so
ὅτι καὶ συντεθέντα ἀνάλογον ἔσται, ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, CF (is) to FD. I say that they will also be proportional


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οὕτως τὸ ΓΔ πρὸς τὸ ΖΔ. (when) composed, (so that) as AB (is) to BE, so CD (is)
Εἰ γὰρ μή ἐστὶν ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ to FD.
πρὸς τὸ ΔΖ, ἔσται ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ For if (it is) not (the case that) as AB is to BE, so
ἤτοι πρὸς ἔλασσόν τι τοῦ ΔΖ ἢ πρὸς μεῖζον. CD (is) to FD, then it will surely be (the case that) as
῎Εστω πρότερον πρὸς ἔλασσον τὸ ΔΗ. καὶ ἐπεί ἐστιν ὡς AB (is) to BE, so CD is either to some (magnitude) less
τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ πρὸς τὸ ΔΗ, συγκείμενα than DF, or (some magnitude) greater (than DF). ‡
μεγέθη ἀνάλογόν ἐστιν· ὥστε καὶ διαιρεθέντα ἀνάλογον Let it, first of all, be to (some magnitude) less (than
ἔσται. ἔστιν ἄρα ὡς τὸ ΑΕ πρὸς τὸ ΕΒ, οὕτως τὸ ΓΗ πρὸς DF), (namely) DG. And since composed magnitudes
τὸ ΗΔ. ὑπόκειται δὲ καὶ ὡς τὸ ΑΕ πρὸς τὸ ΕΒ, οὕτως τὸ are proportional, (so that) as AB is to BE, so CD (is) to
ΓΖ πρὸς τὸ ΖΔ. καὶ ὡς ἄρα τὸ ΓΗ πρὸς τὸ ΗΔ, οὕτως τὸ DG, they will thus also be proportional (when) separated
ΓΖ πρὸς τὸ ΖΔ. μεῖζον δὲ τὸ πρῶτον τὸ ΓΗ τοῦ τρίτου τοῦ [Prop. 5.17]. Thus, as AE is to EB, so CG (is) to GD.
ΓΖ· μεῖζον ἄρα καὶ τὸ δεύτερον τὸ ΗΔ τοῦ τετάρτου τοῦ But it was also assumed that as AE (is) to EB, so CF
ΖΔ. ἀλλὰ καὶ ἔλαττον· ὅπερ ἐστὶν ἀδύνατον· οὐκ ἄρα ἐστὶν (is) to FD. Thus, (it is) also (the case that) as CG (is)
ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ πρὸς ἔλασσον τοῦ to GD, so CF (is) to FD [Prop. 5.11]. And the first
ΖΔ. ὁμοίως δὴ δείξομεν, ὅτι οὐδὲ πρὸς μεῖζον· πρὸς αὐτὸ (magnitude) CG (is) greater than the third CF. Thus,
ἄρα. the second (magnitude) GD (is) also greater than the
᾿Εὰν ἄρα διῃρημένα μεγέθη ἀνάλογον ᾖ, καὶ συντεθέντα fourth FD [Prop. 5.14]. But (it is) also less. The very
ἀνάλογον ἔσται· ὅπερ ἔδει δεῖξαι. thing is impossible. Thus, (it is) not (the case that) as AB
is to BE, so CD (is) to less than FD. Similarly, we can
ijþ they will also be proportional (when) composed. (Which
show that neither (is it the case) to greater (than FD).
Thus, (it is the case) to the same (as FD).
Thus, if separated magnitudes are proportional then

is) the very thing it was required to show.

† In modern notation, this proposition reads that if α : β :: γ : δ then α + β : β :: γ + δ : δ.
‡ Here, Euclid assumes, without proof, that a fourth magnitude proportional to three given magnitudes can always be found.

.
†
Proposition 19
᾿Εὰν ᾖ ὡς ὅλον πρὸς ὅλον, οὕτως ἀφαιρεθὲν πρὸς ἀφαι- If as the whole is to the whole so the (part) taken
ρεθέν, καὶ τὸ λοιπὸν πρὸς τὸ λοιπὸν ἔσται ὡς ὅλον πρὸς away is to the (part) taken away then the remainder to
ὅλον. the remainder will also be as the whole (is) to the whole.
Α Ε Β A E B


Γ Ζ ∆ C F D
῎Εστω γὰρ ὡς ὅλον τὸ ΑΒ πρὸς ὅλον τὸ ΓΔ, οὕτως For let the whole AB be to the whole CD as the (part)
ἀφαιρεθὲν τὸ ΑΕ πρὸς ἀφειρεθὲν τὸ ΓΖ· λέγω, ὅτι καὶ taken away AE (is) to the (part) taken away CF. I say
λοιπὸν τὸ ΕΒ πρὸς λοιπὸν τὸ ΖΔ ἔσται ὡς ὅλον τὸ ΑΒ that the remainder EB to the remainder FD will also be
πρὸς ὅλον τὸ ΓΔ. as the whole AB (is) to the whole CD.
᾿Επεὶ γάρ ἐστιν ὡς τὸ ΑΒ πρὸς τὸ ΓΔ, οὕτως τὸ ΑΕ For since as AB is to CD, so AE (is) to CF, (it is)
πρὸς τὸ ΓΖ, καὶ ἐναλλὰξ ὡς τὸ ΒΑ πρὸς τὸ ΑΕ, οὕτως also (the case), alternately, (that) as BA (is) to AE, so
τὸ ΔΓ πρὸς τὸ ΓΖ. καὶ ἐπεὶ συγκείμενα μεγέθη ἀνάλογόν DC (is) to CF [Prop. 5.16]. And since composed magni-
ἐστιν, καὶ διαιρεθέντα ἀνάλογον ἔσται, ὡς τὸ ΒΕ πρὸς τὸ tudes are proportional then they will also be proportional
ΕΑ, οὕτως τὸ ΔΖ πρὸς τὸ ΓΖ· καὶ ἐναλλάξ, ὡς τὸ ΒΕ πρὸς (when) separated, (so that) as BE (is) to EA, so DF (is)
τὸ ΔΖ, οὕτως τὸ ΕΑ πρὸς τὸ ΖΓ. ὡς δὲ τὸ ΑΕ πρὸς τὸ ΓΖ, to CF [Prop. 5.17]. Also, alternately, as BE (is) to DF,
οὕτως ὑπόκειται ὅλον τὸ ΑΒ πρὸς ὅλον τὸ ΓΔ. καὶ λοιπὸν so EA (is) to FC [Prop. 5.16]. And it was assumed that
ἄρα τὸ ΕΒ πρὸς λοιπὸν τὸ ΖΔ ἔσται ὡς ὅλον τὸ ΑΒ πρὸς as AE (is) to CF, so the whole AB (is) to the whole CD.
ὅλον τὸ ΓΔ. And, thus, as the remainder EB (is) to the remainder
᾿Εὰν ἄρα ᾖ ὡς ὅλον πρὸς ὅλον, οὕτως ἀφαιρεθὲν πρὸς FD, so the whole AB will be to the whole CD.


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ἀφαιρεθέν, καὶ τὸ λοιπὸν πρὸς τὸ λοιπὸν ἔσται ὡς ὅλον Thus, if as the whole is to the whole so the (part)
πρὸς ὅλον [ὅπερ ἔδει δεῖξαι]. taken away is to the (part) taken away then the remain-
[Καὶ ἐπεὶ ἐδείχθη ὡς τὸ ΑΒ πρὸς τὸ ΓΔ, οὕτως τὸ ΕΒ der to the remainder will also be as the whole (is) to
ìri sma as AB (is) to BE, so CD (is) to FD. Thus, composed
πρὸς τὸ ΖΔ, καὶ ἐναλλὰξ ὡς τὸ ΑΒ πρὸς τὸ ΒΕ οὕτως τὸ the whole. [(Which is) the very thing it was required to
ΓΔ πρὸς τὸ ΖΔ, συγκείμενα ἄρα μεγέθη ἀνάλογόν ἐστιν· show.]
ἐδείχθη δὲ ὡς τὸ ΒΑ πρὸς τὸ ΑΕ, οὕτως τὸ ΔΓ πρὸς τὸ [And since it was shown (that) as AB (is) to CD, so
ΓΖ· καί ἐστιν ἀναστρέψαντι]. EB (is) to FD, (it is) also (the case), alternately, (that)


magnitudes are proportional. And it was shown (that)
as BA (is) to AE, so DC (is) to CF. And (the latter) is
converted (from the former).]


.
‡
Corollary
kþ quired to show.
᾿Εκ δὴ τούτου φανερόν, ὅτι ἐὰν συγκείμενα μεγέθη So (it is) clear, from this, that if composed magni-
ἀνάλογον ᾖ, καὶ ἀναστρέψαντι ἀνάλογον ἔσται· ὅπερ ἔδει tudes are proportional then they will also be proportional
δεῖξαι. (when) converted. (Which is) the very thing it was re-

† In modern notation, this proposition reads that if α : β :: γ : δ then α : β :: α − γ : β − δ.
‡ In modern notation, this corollary reads that if α : β :: γ : δ then α : α − β :: γ : γ − δ.

†
.
Proposition 20
᾿Εὰν ᾖ τρία μεγέθη καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος, If there are three magnitudes, and others of equal
σύνδυο λαμβανόμενα καὶ ἐν τῷ αὐτῷ λόγω, δι᾿ ἴσου δὲ τὸ number to them, (being) also in the same ratio taken two
πρῶτον τοῦ τρίτου μεῖζον ᾖ, καὶ τὸ τέταρτον τοῦ ἕκτου by two, and (if), via equality, the first is greater than the
μεῖζον ἔσται, κἂν ἴσον, ἴσον, κἂν ἔλαττον, ἔλαττον. third then the fourth will also be greater than the sixth.
And if (the first is) equal (to the third then the fourth
will also be) equal (to the sixth). And if (the first is) less
(than the third then the fourth will also be) less (than the
sixth).
Α ∆ A D

Β Ε B E
Γ Ζ C F
῎Εστω τρία μεγέθη τὰ Α, Β, Γ, καὶ ἄλλα αὐτοῖς ἴσα τὸ Let A, B, and C be three magnitudes, and D, E, F
πλῆθος τὰ Δ, Ε, Ζ, σύνδυο λαμβανόμενα ἐν τῷ αὐτῷ λόγῳ, other (magnitudes) of equal number to them, (being) in
ὡς μὲν τὸ Α πρὸς τὸ Β, οὕτως τὸ Δ πρὸς τὸ Ε, ὡς δὲ τὸ Β the same ratio taken two by two, (so that) as A (is) to B,
πρὸς τὸ Γ, οὕτως τὸ Ε πρὸς τὸ Ζ, δι᾿ ἴσου δὲ μεῖζον ἔστω so D (is) to E, and as B (is) to C, so E (is) to F. And let
τὸ Α τοῦ Γ· λέγω, ὅτι καὶ τὸ Δ τοῦ Ζ μεῖζον ἔσται, κἂν A be greater than C, via equality. I say that D will also
ἴσον, ἴσον, κἂν ἔλαττον, ἔλαττον. be greater than F. And if (A is) equal (to C then D will
᾿Επεὶ γὰρ μεῖζόν ἐστι τὸ Α τοῦ Γ, ἄλλο δέ τι τὸ Β, τὸ δὲ also be) equal (to F). And if (A is) less (than C then D
μεῖζον πρὸς τὸ αὐτὸ μείζονα λόγον ἔχει ἤπερ τὸ ἔλαττον, will also be) less (than F).
τὸ Α ἄρα πρὸς τὸ Β μείζονα λόγον ἔχει ἤπερ τὸ Γ πρὸς τὸ For since A is greater than C, and B some other (mag-
Β. ἀλλ᾿ ὡς μὲν τὸ Α πρὸς τὸ Β [οὕτως] τὸ Δ πρὸς τὸ Ε, ὡς nitude), and the greater (magnitude) has a greater ratio
δὲ τὸ Γ πρὸς τὸ Β, ἀνάπαλιν οὕτως τὸ Ζ πρὸς τὸ Ε· καὶ τὸ than the lesser to the same (magnitude) [Prop. 5.8], A
Δ ἄρα πρὸς τὸ Ε μείζονα λόγον ἔχει ἤπερ τὸ Ζ πρὸς τὸ Ε. thus has a greater ratio to B than C (has) to B. But as A
τῶν δὲ πρὸς τὸ αὐτὸ λόγον ἐχόντων τὸ μείζονα λόγον ἔχον (is) to B, [so] D (is) to E. And, inversely, as C (is) to B,
μεῖζόν ἐστιν. μεῖζον ἄρα τὸ Δ τοῦ Ζ. ὁμοίως δὴ δείξομεν, so F (is) to E [Prop. 5.7 corr.]. Thus, D also has a greater
ὅτι κἂν ἴσον ᾖ τὸ Α τῷ Γ, ἴσον ἔσται καὶ τὸ Δ τῷ Ζ, κἂν ratio to E than F (has) to E [Prop. 5.13]. And for (mag-


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ἔλαττον, ἔλαττον. nitudes) having a ratio to the same (magnitude), that
᾿Εὰν ἄρα ᾖ τρία μεγέθη καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος, having the greater ratio is greater [Prop. 5.10]. Thus,
σύνδυο λαμβανόμενα καὶ ἐν τῷ αὐτῷ λόγω, δι᾿ ἴσου δὲ τὸ D (is) greater than F. Similarly, we can show that even if
πρῶτον τοῦ τρίτου μεῖζον ᾖ, καὶ τὸ τέταρτον τοῦ ἕκτου A is equal to C then D will also be equal to F, and even
μεῖζον ἔσται, κἂν ἴσον, ἴσον, κἂν ἔλαττον, ἔλαττον· ὅπερ if (A is) less (than C then D will also be) less (than F).
ἔδει δεῖξαι. Thus, if there are three magnitudes, and others of
equal number to them, (being) also in the same ratio
taken two by two, and (if), via equality, the first is greater
kaþ fourth will also be) equal (to the sixth). And (if the first
than the third, then the fourth will also be greater than
the sixth. And if (the first is) equal (to the third then the

is) less (than the third then the fourth will also be) less
(than the sixth). (Which is) the very thing it was required
to show.

† In modern notation, this proposition reads that if α : β :: δ : ǫ and β : γ :: ǫ : ζ then α T γ as δ T ζ.

.
†
Proposition 21
᾿Εὰν ᾖ τρία μεγέθη καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος If there are three magnitudes, and others of equal
σύνδυο λαμβανόμενα καὶ ἐν τῷ αὐτῷ λόγῳ, ᾖ δὲ τετα- number to them, (being) also in the same ratio taken two
ραγμένη αὐτῶν ἡ ἀναλογία, δι᾿ ἴσου δὲ τὸ πρῶτον τοῦ by two, and (if) their proportion (is) perturbed, and (if),
τρίτου μεῖζον ᾖ, καὶ τὸ τέταρτον τοῦ ἕκτου μεῖζον ἔσται, via equality, the first is greater than the third then the
κἂν ἴσον, ἴσον, κἂν ἔλαττον, ἔλαττον. fourth will also be greater than the sixth. And if (the first
is) equal (to the third then the fourth will also be) equal
(to the sixth). And if (the first is) less (than the third then
the fourth will also be) less (than the sixth).
Α ∆ A D
Β Ε B E
Γ Ζ C F
῎Εστω τρία μεγέθη τὰ Α, Β, Γ καὶ ἄλλα αὐτοῖς ἴσα τὸ Let A, B, and C be three magnitudes, and D, E, F
πλῆθος τὰ Δ, Ε, Ζ, σύνδυο λαμβανόμενα καὶ ἐν τῷ αὐτῷ other (magnitudes) of equal number to them, (being) in
λόγῳ, ἔστω δὲ τεταραγμένη αὐτῶν ἡ ἀναλογία, ὡς μὲν τὸ the same ratio taken two by two. And let their proportion
Α πρὸς τὸ Β, οὕτως τὸ Ε πρὸς τὸ Ζ, ὡς δὲ τὸ Β πρὸς τὸ Γ, be perturbed, (so that) as A (is) to B, so E (is) to F, and
οὕτως τὸ Δ πρὸς τὸ Ε, δι᾿ ἴσου δὲ τὸ Α τοῦ Γ μεῖζον ἔστω· as B (is) to C, so D (is) to E. And let A be greater than
λέγω, ὅτι καὶ τὸ Δ τοῦ Ζ μεῖζον ἔσται, κἂν ἴσον, ἴσον, κἂν C, via equality. I say that D will also be greater than F.
ἒλαττον, ἒλαττον. And if (A is) equal (to C then D will also be) equal (to
᾿Επεὶ γὰρ μεῖζόν ἐστι τὸ Α τοῦ Γ, ἄλλο δέ τι τὸ Β, τὸ F). And if (A is) less (than C then D will also be) less
Α ἄρα πρὸς τὸ Β μείζονα λόγον ἔχει ἤπερ τὸ Γ πρὸς τὸ Β. (than F).
ἀλλ᾿ ὡς μὲν τὸ Α πρὸς τὸ Β, οὕτως τὸ Ε πρὸς τὸ Ζ, ὡς For since A is greater than C, and B some other (mag-
δὲ τὸ Γ πρὸς τὸ Β, ἀνάπαλιν οὕτως τὸ Ε πρὸς τὸ Δ. καὶ nitude), A thus has a greater ratio to B than C (has) to
τὸ Ε ἄρα πρὸς τὸ Ζ μείζονα λόγον ἔχει ἤπερ τὸ Ε πρὸς τὸ B [Prop. 5.8]. But as A (is) to B, so E (is) to F. And,
Δ. πρὸς ὃ δὲ τὸ αὐτὸ μείζονα λόγον ἔχει, ἐκεῖνο ἔλασσόν inversely, as C (is) to B, so E (is) to D [Prop. 5.7 corr.].
ἐστιν· ἔλασσον ἄρα ἐστὶ τὸ Ζ τοῦ Δ· μεῖζον ἄρα ἐστὶ τὸ Δ Thus, E also has a greater ratio to F than E (has) to D
τοῦ Ζ. ὁμοίως δὴ δείξομεν, ὅτι κἂν ἴσον ᾖ τὸ Α τῷ Γ, ἴσον [Prop. 5.13]. And that (magnitude) to which the same
ἔσται καὶ τὸ Δ τῷ Ζ, κἂν ἔλαττον, ἔλαττον. (magnitude) has a greater ratio is (the) lesser (magni-
᾿Εὰν ἄρα ᾖ τρία μεγέθη καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος, tude) [Prop. 5.10]. Thus, F is less than D. Thus, D is
σύνδυο λαμβανόμενα καὶ ἐν τῷ αὐτῷ λόγῳ, ᾖ δὲ τετα- greater than F. Similarly, we can show that even if A is
ραγμένη αὐτῶν ἡ ἀναλογία, δι᾿ ἴσου δὲ τὸ πρῶτον τοῦ τρίτου equal to C then D will also be equal to F, and even if (A
μεῖζον ᾖ, καὶ τὸ τέταρτον τοῦ ἕκτου μεῖζον ἔσται, κἂν ἴσον, is) less (than C then D will also be) less (than F).

150