LMNTS OOK 4 onstruction of Rectilinear igures In and round ircles 109
ÎÇÖÓ. efinitions αʹ. Σχῆμα εὐθύγραμμον εἰς σχῆμα εὐθύγραμμον ἐγγράφ- 1. 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. nd, similarly, a (rectilinear) figure is said to be cirτοῦ, 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 circumκύκλου περιφερείας. scribed. δʹ. Σχῆμα δὲ εὐθύγραμμον περὶ κύκλον περιγράφε- 3. rectilinear figure is said to be inscribed in a cirσθαι 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. nd 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.κύκλοςδὲπερὶσχῆμαπεριγράφεσθαιλέγεται,ὅταν 5. nd, similarly, a circle is said to be inscribed in a 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.ὐθεῖαεἰςκύκλονἐναρμόζεσθαιλέγεται,ὅταντὰ 6. nd 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 touches each angle of the (figure) about which it is circumscribed. 7. straight-line is said to be inserted into a circle when its extemities are on the circumference of the circle.. Proposition 1 ἰς 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. στωὁδοθεὶςκύκλοςὁ,ἡδὲδοθεῖσαεὐθεῖαμὴ Let be the given circle, and the given straightμείζωντῆςτοῦκύκλουδιαμέτρουἡδ.δεῖδὴεἰςτὸν line (which is) not greater than the diameter of the cirκύκλον τῇ Δ εὐθείᾳ ἴσην εὐθεῖαν ἐναρμόσαι. cle. So it is required to insert a straight-line, equal to the Ηχθω τοῦ κύκλου διάμετρος ἡ. εἰ μὲν οὖν ἴση straight-line, into the circle. ἐστὶνἡτῇδ,γεγονὸςἂνεἴητὸἐπιταχθέν ἐνήρμοσται Let a diameter of circle have been drawn. 110
γὰρεἰςτὸνκύκλοντῇδεὐθείᾳἴσηἡ.εἰδὲμείζων Therefore, if is equal to then that (which) was ἐστὶνἡτῆςδ,κείσθωτῇδἴσηἡ,καὶκέντρῳ prescribed has taken place. or the (straight-line), τῷδιαστήματιδὲτῷκύκλοςγεγράφθωὁ,καὶ equal to the straight-line, has been inserted into the ἐπεζεύχθω ἡ. circle. nd if is greater than then let be πεὶοὖντοσημεῖονκέντρονἐστὶτοῦκύκλου, made equal to [Prop. 1.3], and let the circle have ἴσηἐστὶνἡτῇ.ἀλλὰτῇδἡἐστινἴση καὶἡδ been drawn with center and radius. nd let ἄρα τῇ ἐστιν ἴση. have been joined. ἰςἄρατὸνδοθέντακύκλοντὸντῇδοθείσῃεὐθείᾳ Therefore, since the point is the center of circle τῇδἴσηἐνήρμοσταιἡ ὅπερἔδειποιῆσαι., is equal to. ut, is equal to. Thus, is also equal to. Thus,, equal to the given straight-line, has been inserted into the given circle. (Which is) the very 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 ἰς τὸν δοθέντα κύκλον τῷ δοθέντι τριγώνῳ ἰσογώνιον To inscribe a triangle, equiangular with a given trianτρίγωνονἐγγράψαι. gle, in a given circle. Η G Θ H στωὁδοθεὶςκύκλοςὁ,τὸδὲδοθὲντριγωνον Let be the given circle, and the given triτὸ Δ δεῖ δὴ εἰς τὸν κύκλον τῷ Δ τριγώνῳ angle. So it is required to inscribe a triangle, equiangular ἰσογώνιοντρίγωνονἐγγράψαι. with triangle, in circle. ΗχθωτοῦκύκλουἐφαπτομένηἡΗΘκατὰτὸ, Let GH have been drawn touching circle at. καὶσυνεστάτωπρὸςτῇθεὐθείᾳκαὶτῷπρὸςαὐτῇσημείῳ nd let (angle) H, equal to angle, have been τῷτῇὑπὸδγωνίᾳἴσηἡὑπὸθ,πρὸςδὲτῇη constructed on the straight-line H at the point on it, εὐθείᾳκαὶτῷπρὸςαὐτῇσημείῳτῷτῇὑπὸδ[γωνίᾳ] and (angle) G, equal to [angle], on the straight- ἴσηἡὑπὸη,καὶἐπεζεύχθωἡ. line G at the point on it [Prop. 1.23]. nd let πεὶ οὖν κύκλου τοῦ ἐφάπτεταί τις εὐθεῖα ἡ Θ, have been joined. καὶ ἀπὸ τῆς κατὰ τὸ ἐπαφῆς εἰς τὸν κύκλον διῆκται εὐθεῖα Therefore, since some straight-line H touches the ἡ,ἡἄραὑπὸθἴσηἐστὶτῇἐντῷἐναλλὰξτοῦκύκλου circle, and the straight-line has been drawn τμήματιγωνίᾳτῇὑπὸ.ἀλλ ἡὑπὸθτῇὑπὸδ across (the circle) from the point of contact, (angle) ἐστινἴση καὶἡὑπὸἄραγωνίατῇὑπὸδἐστιν H is thus equal to the angle in the alternate ἴση. διὰτὰαὐτὰδὴκαὶἡὑπὸτῇὑπὸδἐστιν segment of the circle [Prop. 3.32]. ut, H is equal to ἴση καὶλοιπὴἄραἡὑπὸλοιπῇτῇὑπὸδἐστινἴση. Thus, angle is also equal to. So, for the [ἰσογώνιονἄραἐστὶτὸτρίγωνοντῷδτριγώνῳ, same (reasons), is also equal to. Thus, the reκαὶ ἐγγέγραπται εἰς τὸν κύκλον]. maining (angle) is equal to the remaining (angle) ἰς τὸν δοθέντα ἄρα κύκλον τῷ δοθέντι τριγώνῳ [Prop. 1.32]. [Thus, triangle is equiangu- ἰσογώνιον τρίγωνον ἐγγέγραπται ὅπερ ἔδει ποιῆσαι. lar with triangle, and has been inscribed in circle 111
See the footnote to Prop. 3.34. ]. 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.. Proposition 3 Περὶ τὸν δοθέντα κύκλον τῷ δοθέντι τριγώνῳ ἰσογώνιον To circumscribe a triangle, equiangular with a given τρίγωνονπεριγράψαι. triangle, about a given circle. Λ Μ Κ Θ Ν Η M K H L N G στωὁδοθεὶςκύκλοςὁ,τὸδὲδοθὲντρίγωνον Let be the given circle, and the given triτὸδ δεῖδὴπερὶτὸνκύκλοντῷδτριγώνῳ angle. So it is required to circumscribe a triangle, equian- ἰσογώνιον τρίγωνον περιγράψαι. gular with triangle, about circle. κβεβλήσθω ἡ ἐφ ἑκάτερα τὰ μέρη κατὰ τὰ Η, Θ Let have been produced in each direction to σημεῖα,καὶεἰλήφθωτοῦκύκλουκέντροντὸκ,καὶ points G and H. nd let the center K of circle διήχθω, ὡς ἔτυχεν, εὐθεῖα ἡ Κ, καὶ συνεστάτω πρὸς τῇ have been found [Prop. 3.1]. nd let the straight-line ΚεὐθείᾳκαὶτῷπρὸςαὐτῇσημείῳτῷΚτῇμὲνὑπὸΔΗ K have been drawn, at random, across (). nd γωνίᾳἴσηἡὑπὸκ,τῇδὲὑπὸδθἴσηἡὑπὸκ,καὶ let (angle) K, equal to angle G, have been conδιὰτῶν,,σημείωνἤχθωσανἐφαπτόμεναιτοῦ structed on the straight-line K at the point K on it, κύκλουαἱλμ,μν,νλ. and (angle) K, equal to H [Prop. 1.23]. nd let ΚαὶἐπεὶἐφάπτονταιτοῦκύκλουαἱΛΜ,ΜΝ,ΝΛ the (straight-lines) LM, MN, and NL have been κατὰτὰ,,σημεῖα,ἀπὸδὲτοῦκκέντρουἐπὶτὰ,, drawn through the points,, and (respectively), σημεῖαἐπεζευγμέναιεἰσὶναἱκ,κ,κ,ὀρθαὶἄραεἰσὶν touching the circle. αἱπρὸςτοῖς,,σημείοιςγωνίαι.καὶἐπεὶτοῦμκ nd since LM, MN, and NL touch circle at τετραπλεύρου αἱ τέσσαρες γωνίαι τέτρασιν ὀρθαῖς ἴσαι εἰσίν, points,, and (respectively), and K, K, and ἐπειδήπερκαὶεἰςδύοτρίγωναδιαιρεῖταιτὸμκ,καίεἰσιν K are joined from the center K to points,, and ὀρθαὶαἱὑπὸκμ,κμγωνίαι,λοιπαὶἄρααἱὑπὸκ, (respectively), the angles at points,, and are Μδυσὶνὀρθαῖςἴσαιεἰσίν. εἰσὶδὲκαὶαἱὑπὸδη, thus right-angles [Prop. 3.18]. nd since the (sum of the) Δδυσὶνὀρθαῖςἴσαι αἱἄραὑπὸκ,μταῖςὑπὸ four angles of quadrilateral MK is equal to four right- ΔΗ, Δ ἴσαι εἰσίν, ὧν ἡ ὑπὸ Κ τῇ ὑπὸ ΔΗ ἐστιν angles, inasmuch as M K (can) also (be) divided into ἴση λοιπὴἄραἡὑπὸμλοιπῇτῇὑπὸδἐστινἴση. two triangles [Prop. 1.32], and angles KM and KM ὁμοίωςδὴδειχθήσεται,ὅτικαὶἡὑπὸλντῇὑπὸδ are (both) right-angles, the (sum of the) remaining (an- ἐστιν ἴση καὶ λοιπὴ ἄρα ἡ ὑπὸ ΜΛΝ[λοιπῇ] τῇ ὑπὸ Δ gles), K and M, is thus equal to two right-angles. ἐστινἴση.ἰσογώνιονἄραἐστὶτὸλμντρίγωνοντῷδ nd G and is also equal to two right-angles τριγώνῳ καὶ περιγέγραπται περὶ τὸν κύκλον. [Prop. 1.13]. Thus, K and M is equal to G Περὶ τὸν δοθέντα ἄρα κύκλον τῷ δοθέντι τριγώνῳ and, of which K is equal to G. Thus, the re- ἰσογώνιον τρίγωνον περιγέγραπται ὅπερ ἔδει ποιῆσαι. mainder M is equal to the remainder. So, similarly, it can be shown that LN is also equal to. Thus, the remaining (angle) MLN is also equal to the 112
See the footnote to Prop. 3.34. [remaining] (angle) [Prop. 1.32]. Thus, triangle LMN is equiangular with triangle. nd it has been drawn around circle. 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.. Proposition 4 ἰςτὸδοθὲντρίγωνονκύκλονἐγγράψαι. To inscribe a circle in a given triangle. Η G στωτὸδοθὲντρίγωνοντὸ δεῖδὴεἰςτὸ Let be the given triangle. So it is required to τρίγωνον κύκλον ἐγγράψαι. inscribe a circle in triangle. Τετμήσθωσαναἱὑπὸ,γωνίαιδίχαταῖςΔ, Let the angles and have been cut in half by Δ εὐθείαις, καὶ συμβαλλέτωσαν ἀλλήλαις κατὰ τὸ Δ the straight-lines and (respectively) [Prop. 1.9], σημεῖον, καὶ ἤχθωσαν ἀπὸ τοῦ Δ ἐπὶ τὰς,, and let them meet one another at point, and let, εὐθείαςκάθετοιαἱδ,δ,δη., and G have been drawn from point, perpendic- Καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ Δ γωνία τῇ ὑπὸ Δ, ular to the straight-lines,, and (respectively) ἐστὶδὲκαὶὀρθὴἡὑπὸδὀρθῇτῇὑπὸδἴση,δύο [Prop. 1.12]. δὴτρίγωνάἐστιτὰδ,δτὰςδύογωνίαςταῖςδυσὶ nd since angle is equal to, and the rightγωνίαιςἴσαςἔχοντακαὶμίανπλευρὰνμιᾷπλευρᾷἴσηντὴν angle is also equal to the right-angle, ὑποτείνουσαν ὑπὸ μίαν τῶν ἴσων γωνιῶν κοινὴν αὐτῶν τὴν and are thus two triangles having two angles equal Δ καὶ τὰς λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας to two angles, and one side equal to one side the (one) ἕξουσιν ἴσηἄραἡδτῇδ.διὰτὰαὐτὰδὴκαὶἡδη subtending one of the equal angles (which is) common to τῇ Δ ἐστιν ἴση. αἱ τρεῖς ἄρα εὐθεῖαι αἱ Δ, Δ, ΔΗ the (triangles) (namely),. Thus, they will also have ἴσαι ἀλλήλαις εἰσίν ὁ ἄρα κέντρῷ τῷ Δ καὶ διαστήματι ἑνὶ the remaining sides equal to the (corresponding) remainτῶν,,ηκύκλοςγραφόμενοςἥξεικαὶδιὰτῶνλοιπῶν ing sides [Prop. 1.26]. Thus, (is) equal to. So, σημείων καὶ ἐφάψεται τῶν,, εὐθειῶν διὰ τὸ for the same (reasons), G is also equal to. Thus, ὀρθὰςεἶναιτὰςπρὸςτοῖς,,ησημείοιςγωνίας. εἰγὰρ the three straight-lines,, and G are equal to τεμεῖαὐτάς,ἔσταιἡτῇδιαμέτρῳτοῦκύκλουπρὸςὀρθὰς one another. Thus, the circle drawn with center, and ἀπ ἄκραςἀγομένηἐντὸςπίπτουσατοῦκύκλου ὅπερἄτο- radius one of,, or G, will also go through the reπονἐδείχθη οὐκἄραὁκέντρῳτῷδδιαστήματιδὲἑνὶτῶν maining points, and will touch the straight-lines,,,,ηγραφόμενοςκύκλοςτεμεῖτὰς,,εὐθείας and, on account of the angles at,, and G being ἐφάψεται ἄρα αὐτῶν, καὶ ἔσται ὁ κύκλος ἐγγεγραμμένος εἰς right-angles. or if it cuts (one of) them then it will be τὸ τρίγωνον. ἐγγεγράφθω ὡς ὁ Η. a (straight-line) drawn at right-angles to a diameter of ἰς ἄρα τὸ δοθὲν τρίγωνον τὸ κύκλος ἐγγέγραπται the circle, from its extremity, falling inside the circle. The ὁη ὅπερἔδειποιῆσαι. very thing was shown (to be) absurd [Prop. 3.16]. Thus, the circle drawn with center, and radius one of,, 113
Here, and in the following propositions, it is understood that the radius is actually one of,, or G. or G, does not cut the straight-lines,, and. Thus, it will touch them and will be the circle inscribed in triangle. Let it have been (so) inscribed, like G (in the figure). Thus, the circle G has been inscribed in the given triangle. (Which is) the very thing it was required to do.. Proposition 5 Περὶτὸδοθὲντρίγωνονκύκλονπεριγράψαι. To circumscribe a circle about a given triangle. στωτὸδοθὲντρίγωνοντὸ δεῖδὲπερὶτὸδοθὲν Let be the given triangle. So it is required to τρίγωνοντὸκύκλονπεριγράψαι. circumscribe a circle about the given triangle. Τετμήσθωσαναἱ,εὐθεῖαιδίχακατὰτὰΔ, Let the straight-lines and have been cut in σημεῖα,καὶἀπὸτῶνδ,σημείωνταῖς,πρὸςὀρθὰς half at points and (respectively) [Prop. 1.10]. nd ἤχθωσαναἱδ, συμπεσοῦνταιδὴἤτοιἐντὸςτοῦ let and have been drawn from points and, τριγώνουἢἐπὶτῆςεὐθείαςἢἐκτὸςτῆς. at right-angles to and (respectively) [Prop. 1.11]. Συμπιπτέτωσαν πρότερον ἐντὸς κατὰ τὸ, καὶ ἐπεζεύχθ- So ( and ) will surely either meet inside triangle ωσαν αἱ,,. καὶ ἐπεὶ ἴση ἐστὶν ἡ Δ τῇ Δ, κοινὴ, on the straight-line, or beyond. δὲκαὶπρὸςὀρθὰςἡδ,βάσιςἄραἡβάσειτῇἐστιν Let them, first of all, meet inside (triangle ) at ἴση. ὁμοίωςδὴδείξομεν,ὅτικαὶἡτῇἐστινἴση (point), and let,, and have been joined. ὥστεκαὶἡτῇἐστινἴση αἱτρεῖςἄρααἱ,, nd since is equal to, and is common and ἴσαι ἀλλήλαις εἰσίν. ὁ ἄρα κέντρῳ τῷ διαστήματι δὲ ἑνὶ at right-angles, the base is thus equal to the base τῶν,,κύκλοςγραφόμενοςἥξεικαὶδιὰτῶνλοιπῶν [Prop. 1.4]. So, similarly, we can show that is also σημείων,καὶἔσταιπεριγεγραμμένοςὁκύκλοςπερὶτὸ equal to. So that is also equal to. Thus, the τρίγωνον.περιγεγράφθωὡςὁ. three (straight-lines),, and are equal to one ἈλλὰδὴαἱΔ,συμπιπτέτωσανἐπὶτῆςεὐθείας another. Thus, the circle drawn with center, and radius κατὰ τὸ, ὡς ἔχει ἐπὶ τῆς δευτέρας καταγραφῆς, καὶ one of,, or, will also go through the remaining ἐπεζεύχθω ἡ. ὁμοίως δὴ δείξομεν, ὅτι τὸ σημεῖον points. nd the circle will have been circumscribed about κέντρον ἐστὶ τοῦ περὶ τὸ τρίγωνον περιγραφομένου triangle. Let it have been (so) circumscribed, like κύκλου. (in the first diagram from the left). ἈλλὰδὴαἱΔ,συμπιπτέτωσανἐκτὸςτοῦ nd so, let and meet on the straight-line τριγώνουκατὰτὸπάλιν,ὡςἔχειἐπὶτῆςτρίτηςκατα- at (point), like in the second diagram (from the γραφῆς, καί ἐπεζεύχθωσαν αἱ,,. καὶ ἐπεὶ πάλιν left). nd let have been joined. So, similarly, we can ἴσηἐστὶνἡδτῇδ,κοινὴδὲκαὶπρὸςὀρθὰςἡδ,βάσις show that point is the center of the circle circumscribed ἄρα ἡ βάσει τῇ ἐστιν ἴση. ὁμοίως δὴ δείξομεν, ὅτι about triangle. καὶἡτῇἐστινἴση ὥστεκαὶἡτῇἐστινἴση nd so, let and meet outside triangle, ὁἄρα[πάλιν]κέντρῳτῷδιαστήματιδὲἑνὶτῶν,, again at (point), like in the third diagram (from the κύκλοςγραφόμενοςἥξεικαὶδιὰτῶνλοιπῶνσημείων, left). nd let,, and have been joined. nd, καὶ ἔσται περιγεγραμμένος περὶ τὸ τρίγωνον. again, since is equal to, and is common and Περὶ τὸ δοθὲν ἄρα τρίγωνον κύκλος περιγέγραπται at right-angles, the base is thus equal to the base ὅπερ ἔδει ποιῆσαι. [Prop. 1.4]. So, similarly, we can show that is also equal to. So that is also equal to. Thus, 114
[again] the circle drawn with center, and radius one of,, and, will also go through the remaining points. nd it will have been circumscribed about triangle. Thus, a circle has been circumscribed about the given triangle. (Which is) the very thing it was required to do.. Proposition 6 ἰςτὸνδοθέντακύκλοντετράγωνονἐγγράψαι. To inscribe a square in a given circle. στωἡδοθεὶςκύκλοςὁδ δεῖδὴεἰςτὸνδ Let be the given circle. So it is required to κύκλοντετράγωνονἐγγράψαι. inscribe a square in circle. Ηχθωσαν τοῦ Δ κύκλου δύο διάμετροι πρὸς ὀρθὰς Let two diameters of circle, and, have ἀλλήλαιςαἱ,δ,καὶἐπεζεύχθωσαναἱ,,δ, been drawn at right-angles to one another. nd let, Δ.,, and have been joined. ΚαὶἐπεὶἴσηἐστὶνἡτῇΔ κέντρονγὰρτὸ κοινὴ nd since is equal to, for (is) the center δὲκαὶπρὸςὀρθὰςἡ,βάσιςἄραἡβάσειτῇδἴση (of the circle), and is common and at right-angles, ἐστίν. διὰτὰαὐτὰδὴκαὶἑκατέρατῶν,δἑκατέρᾳ the base is thus equal to the base [Prop. 1.4]. τῶν,δἴσηἐστίν ἰσόπλευρονἄραἐστὶτὸδ So, for the same (reasons), each of and is equal τετράπλευρον. λέγω δή, ὅτι καὶ ὀρθογώνιον. ἐπεὶ γὰρ ἡ to each of and. Thus, the quadrilateral Δ εὐθεῖα διάμετρός ἐστι τοῦ Δ κύκλου, ἡμικύκλιον is equilateral. So I say that (it is) also right-angled. or ἄρα ἐστὶ τὸ Δ ὀρθὴ ἄρα ἡ ὑπὸ Δ γωνία. διὰ τὰ since the straight-line is a diameter of circle, αὐτὰ δὴ καὶ ἑκάστη τῶν ὑπὸ, Δ, Δ ὀρθή ἐστιν is thus a semi-circle. Thus, angle (is) a right- ὀρθογώνιον ἄρα ἐστὶ τὸ Δ τετράπλευρον. ἐδείχθη δὲ angle [Prop. 3.31]. So, for the same (reasons), (angles) καὶ ἰσόπλευρον τετράγωνον ἄρα ἐστίν. καὶ ἐγγέγραπται εἰς,, and are also each right-angles. Thus, τὸν Δ κύκλον. the quadrilateral is right-angled. nd it was also ἰς ἄρα τὸν δοθέντα κύκλον τετράγωνον ἐγγέγραπται shown (to be) equilateral. Thus, it is a square [ef. 1.22]. τὸδ ὅπερἔδειποιῆσαι. nd it has been inscribed in circle. Thus, the square has been inscribed in the given circle. (Which is) the very thing it was required 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]. Þ. Proposition 7 Περὶτὸνδοθέντακύκλοντετράγωνονπεριγράψαι. To circumscribe a square about a given circle. 115
στωὁδοθεὶςκύκλοςὁδ δεῖδὴπερὶτὸνδ Let be the given circle. So it is required to κύκλοντετράγωνονπεριγράψαι. circumscribe a square about circle. Η G Θ Κ H K Ηχθωσαν τοῦ Δ κύκλου δύο διάμετροι πρὸς ὀρθὰς Let two diameters of circle, and, have ἀλλήλαιςαἱ,δ,καὶδιὰτῶν,,,δσημείωνἤχθω- been drawn at right-angles to one another. nd let G, σανἐφαπτόμεναιτοῦδκύκλουαἱη,ηθ,θκ,κ. GH, HK, and K have been drawn through points, πεὶοὖνἐφάπτεταιἡητοῦδκύκλου,ἀπὸδὲ,, and (respectively), touching circle. τοῦκέντρουἐπὶτὴνκατὰτὸἐπαφὴνἐπέζευκταιἡ Therefore, since G touches circle, and, αἱ ἄρα πρὸς τῷ γωνίαι ὀρθαί εἰσιν. διὰ τὰ αὐτὰ has been joined from the center to the point of contact δὴ καὶ αἱ πρὸς τοῖς,, Δ σημείοις γωνίαι ὀρθαί εἰσιν., the angles at are thus right-angles [Prop. 3.18]. So, καὶἐπεὶὀρθήἐστινἡὑπὸγωνία,ἐστὶδὲὀρθὴκαὶἡ for the same (reasons), the angles at points,, and ὑπὸη,παράλληλοςἄραἐστὶνἡηθτῇ.διὰτὰαὐτὰ are also right-angles. nd since angle is a rightδὴκαὶἡτῇκἐστιπαράλληλος. ὥστεκαὶἡηθτῇ angle, and G is also a right-angle, GH is thus parallel Κ ἐστι παράλληλος. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἑκατέρα to [Prop. 1.29]. So, for the same (reasons), is τῶνη,θκτῇδἐστιπαράλληλος.παραλληλόγραμμα also parallel to K. So that GH is also parallel to K ἄραἐστὶτὰηκ,η,κ,,κ ἴσηἄραἐστὶνἡμὲν [Prop. 1.30]. So, similarly, we can show that G and ΗτῇΘΚ,ἡδὲΗΘτῇΚ.καὶἐπεὶἴσηἐστὶνἡτῇ HK are each parallel to. Thus, GK, G, K,, Δ,ἀλλὰκαὶἡμὲνἑκατέρᾳτῶνΗΘ,Κ,ἡδὲΔ and K are (all) parallelograms. Thus, G is equal to ἑκατέρᾳτῶνη,θκἐστινἴση[καὶἑκατέραἄρατῶνηθ, HK, and GH to K [Prop. 1.34]. nd since is equal ΚἑκατέρᾳτῶνΗ,ΘΚἐστινἴση],ἰσόπλευρονἄραἐστὶτὸ to, but (is) also (equal) to each of GH and K, ΗΘΚτετράπλευρον. λέγωδή,ὅτικαὶὀρθογώνιον. ἐπεὶ and is equal to each of G and HK [Prop. 1.34] γὰρπαραλληλόγραμμόνἐστιτὸη,καίἐστινὀρθὴἡ [and each of GH and K is thus equal to each of G ὑπὸ,ὀρθὴἄρακαὶἡὑπὸη.ὁμοίωςδὴδείξομεν, and HK], the quadrilateral GHK is thus equilateral. ὅτικαὶαἱπρὸςτοῖςθ,κ,γωνίαιὀρθαίεἰσιν.ὀρθογώνιον So I say that (it is) also right-angled. or since G ἄρα ἐστὶ τὸ ΗΘΚ. ἐδείχθη δὲ καὶ ἰσόπλευρον τετράγωνον is a parallelogram, and is a right-angle, G is ἄρα ἐστίν. καὶ περιγέγραπται περὶ τὸν Δ 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 are also right-angles. ὅπερ ἔδει ποιῆσαι. Thus, GHK is right-angled. nd it was also shown (to be) equilateral. Thus, it is a square [ef. 1.22]. nd it has been circumscribed about circle. 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. 116
. Proposition 8 ἰςτὸδοθὲντετράγωνονκύκλονἐγγράψαι. To inscribe a circle in a given square. στωτὸδοθὲντετράγωνοντὸδ.δεῖδὴεἰςτὸ Let the given square be. So it is required to Δτετράγωνονκύκλονἐγγράψαι. inscribe a circle in square. Η Κ G K Θ H ΤετμήσθωἑκατέρατῶνΔ,δίχακατὰτὰ, Let and each have been cut in half at points σημεῖα, καὶ διὰ μὲν τοῦ ὁποτέρᾳ τῶν, Δ παράλληλος and (respectively) [Prop. 1.10]. nd let H have been ἤχθωὁθ,διὰδὲτοῦὁποτέρᾳτῶνδ,παράλληλος drawn through, parallel to either of or, and let ἤχθω ἡ Κ παραλληλόγραμμονἄρα ἐστὶν ἕκαστον τῶν K have been drawn through, parallel to either of Κ, Κ, Θ, ΘΔ, Η, Η, Η, ΗΔ, καὶ αἱ ἀπεναντίον or [Prop. 1.31]. Thus, K, K, H, H, G, G, αὐτῶν πλευραὶ δηλονότι ἴσαι[εἰσίν]. καὶ ἐπεὶ ἴση ἐστὶν ἡ G, and G are each parallelograms, and their opposite Δτῇ,καίἐστιτῆςμὲνΔἡμίσειαἡ,τῆςδὲ sides [are] manifestly equal [Prop. 1.34]. nd since ἡμίσειαἡ,ἴσηἄρακαὶἡτῇ ὥστεκαὶαἱἀπε- is equal to, and is half of, and half of ναντίον ἴσηἄρακαὶἡητῇη.ὁμοίωςδὴδείξομεν,ὅτι, (is) thus also equal to. So that the opposite καὶ ἑκατέρα τῶν ΗΘ, ΗΚ ἑκατέρᾳ τῶν Η, Η ἐστιν ἴση (sides are) also (equal). Thus, G (is) also equal to G. αἱ τέσσαρες ἄρα αἱ Η, Η, 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,, Θ, Κ is equal to each of G and G. Thus, the four (straightκύκλοςγραφόμενοςἥξεικαὶδιὰτῶνλοιπῶνσημείων καὶ lines) G, G, GH, and GK [are] equal to one another. ἐφάψεταιτῶν,,δ,δεὐθειῶνδιὰτὸὀρθὰςεἶναι Thus, the circle drawn with center G, and radius one of τὰςπρὸςτοῖς,,θ,κγωνίας εἰγὰρτεμεῖὁκύκλοςτὰς,, H, or K, will also go through the remaining points.,,δ,δ,ἡτῇδιαμέτρῳτοῦκύκλουπρὸςὀρθὰς nd it will touch the straight-lines,,, and ἀπ ἄκραςἀγομένηἐντὸςπεσεῖταιτοῦκύκλου ὅπερἄτοπον, on account of the angles at,, H, and K being ἐδείχθη.οὐκἄραὁκέντρῳτῷηδιαστήματιδὲἑνὶτῶν, right-angles. or if the circle cuts,,, or,, Θ, Κ κύκλος γραφόμενος τεμεῖ τὰς,, Δ, Δ then a (straight-line) drawn at right-angles to a diameter εὐθείας. ἐφάψεται ἄρα αὐτῶν καὶ ἔσται ἐγγεγραμμένος εἰς of the circle, from its extremity, will fall inside the circle. τὸ Δ τετράγωνον. The very thing was shown (to be) absurd [Prop. 3.16]. ἰς ἄρα τὸ δοθὲν τετράγωνον κύκλος ἐγγέγραπται Thus, the circle drawn with center G, and radius one of ὅπερἔδειποιῆσαι.,, H, or K, does not cut the straight-lines,,, or. Thus, it will touch them, and will have been inscribed in the square. Thus, a circle has been inscribed in the given square. (Which is) the very thing it was required to do.. Proposition 9 Περὶτὸδοθὲντετράγωνονκύκλονπεριγράψαι. To circumscribe a circle about a given square. στωτὸδοθὲντετράγωνοντὸδ δεῖδὴπερὶτὸ Let be the given square. So it is required to Δτετράγωνονκύκλονπεριγράψαι. circumscribe a circle about square. 117
πιζευχθεῖσαι γὰρ αἱ, Δ τεμνέτωσαν ἀλλήλας and being joined, let them cut one another at κατὰτὸ.. ΚαὶἐπεὶἴσηἐστὶνἡΔτῇ,κοινὴδὲἡ,δύο nd since is equal to, and (is) common, δὴαἱδ,δυσὶταῖς,ἴσαιεἰσίν καὶβάσιςἡ the two (straight-lines), are thus equal to the two Δβάσειτῇἴση γωνίαἄραἡὑπὸδγωνίᾳτῇὑπὸ (straight-lines),. nd the base (is) equal to ἴσηἐστίν ἡἄραὑπὸδγωνίαδίχατέτμηταιὑπὸ the base. Thus, angle is equal to angle τῆς.ὁμοίωςδὴδείξομεν,ὅτικαὶἑκάστητῶνὑπὸ, [Prop. 1.8]. Thus, the angle has been cut in half Δ, Δ δίχα τέτμηται ὑπὸ τῶν, Δ εὐθειῶν. καὶ by. So, similarly, we can show that,, and ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ Δ γωνία τῇ ὑπὸ, καί ἐστι have each been cut in half by the straight-lines τῆςμὲνὑπὸδἡμίσειαἡὑπὸ,τῆςδὲὑπὸ and. nd since angle is equal to, and ἡμίσειαἡὑπὸ,καὶἡὑπὸἄρατῇὑπὸἐστιν is half of, and half of, is ἴση ὥστεκαὶπλευρὰἡτῇἐστινἴση. ὁμοίωςδὴ thus also equal to. So that side is also equal δείξομεν, ὅτι καὶ ἑκατέρα τῶν, [εὐθειῶν] ἑκατέρᾳ to [Prop. 1.6]. So, similarly, we can show that each τῶν, Δ ἴση ἐστίν. αἱ τέσσαρες ἄρα αἱ,,, of the [straight-lines] and are also equal to each Δἴσαιἀλλήλαιςεἰσίν.ὁἄρακέντρῳτῷκαὶδιαστήματι of and. Thus, the four (straight-lines),, ἑνὶτῶν,,,δκύκλοςγραφόμενοςἥξεικαὶδιὰτῶν, and are equal to one another. Thus, the circle λοιπῶνσημείωνκαὶἔσταιπεριγεγραμμένοςπερὶτὸδ drawn with center, and radius one of,,, or, τετράγωνον. περιγεγράφθω ὡς ὁ Δ. will also go through the remaining points, and will have Περὶ τὸ δοθὲν ἄρα τετράγωνον κύκλος περιγέγραπται been circumscribed about the square. Let it have ὅπερἔδειποιῆσαι. been (so) circumscribed, like (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 πρὸςτῇβάσειγωνιῶνδιπλασίονατῆςλοιπῆς. angles at the base double the remaining (angle). κκείσθω τις εὐθεῖα ἡ, καὶ τετμήσθω κατὰ τὸ Let some straight-line be taken, and let it have σημεῖον, ὥστε τὸ ὑπὸ τῶν, περιεχόμενον been cut at point so that the rectangle contained by ὀρθογώνιον ἴσον εἶναι τῷ ἀπὸ τῆς τετραγώνῳ καὶ and is equal to the square on [Prop. 2.11]. κέντρῳ τῷ καὶ διαστήματι τῷ κύκλος γεγράφθω nd let the circle have been drawn with center, ὁδ,καὶἐνηρμόσθωεἰςτὸνδκύκλοντῇεὐθείᾳ and radius. nd let the straight-line, equal to μὴ μείζονι οὔσῃ τῆς τοῦ Δ κύκλου διαμέτρου ἴση εὐθεῖα the straight-line, being not greater than the diame- ἡ Δ καὶ ἐπεζεύχθωσαν αἱ Δ, Δ, καὶ περιγεγράφθω ter of circle, have been inserted into circle περὶτὸδτρίγωνονκύκλοςὁδ. [Prop. 4.1]. nd let and have been joined. nd let the circle have been circumscribed about triangle [Prop. 4.5]. 118
Καὶἐπεὶτὸὑπὸτῶν,ἴσονἐστὶτῷἀπὸτῆς nd since the (rectangle contained) by and,ἴσηδὲἡτῇδ,τὸἄραὑπὸτῶν,ἴσον is equal to the (square) on, and (is) equal to ἐστὶτῷἀπὸτῆςδ.καὶἐπεὶκύκλουτοῦδεἴληπταίτι, the (rectangle contained) by and is thus σημεῖονἐκτὸςτὸ,καὶἀπὸτοῦπρὸςτὸνδκύκλον equal to the (square) on. nd since some point προσπεπτώκασιδύοεὐθεῖαιαἱ,δ,καὶἡμὲναὐτῶν has been taken outside of circle, and two straightτέμνει,ἡδὲπροσπίπτει,καίἐστιτὸὑπὸτῶν,ἴσον lines and have radiated from towards the cirτῷἀπὸτῆςδ,ἡδἄραἐφάπτεταιτοῦδκύκλου.ἐπεὶ cle, and (one) of them cuts (the circle), and (the οὖνἐφάπτεταιμὲνἡδ,ἀπὸδὲτῆςκατὰτὸδἐπαφῆς other) meets (the circle), and the (rectangle contained) διῆκται ἡ Δ, ἡ ἄρα ὑπὸ Δ γωνιά ἴση ἐστὶ τῇ ἐν τῷ by and is equal to the (square) on, thus ἐναλλὰξ τοῦ κύκλου τμήματι γωνίᾳ τῇ ὑπὸ Δ. ἐπεὶ οὖν touches circle [Prop. 3.37]. Therefore, since ἴση ἐστὶν ἡ ὑπὸ Δ τῇ ὑπὸ Δ, κοινὴ προσκείσθω ἡ touches (the circle), and has been drawn across (the ὑπὸδ ὅληἄραἡὑπὸδἴσηἐστὶδυσὶταῖςὑπὸδ, circle) from the point of contact, the angle is Δ. ἀλλὰ ταῖς ὑπὸ Δ, Δ ἴση ἐστὶν ἡ ἐκτὸς ἡ ὑπὸ thus equal to the angle in the alternate segment of Δ καὶἡὑπὸδἄραἴσηἐστὶτῇὑπὸδ.ἀλλὰἡ the circle [Prop. 3.32]. Therefore, since is equal ὑπὸδτῇὑπὸδἐστινἴση,ἐπεὶκαὶπλευρὰἡδ to, let have been added to both. Thus, the τῇἐστινἴση ὥστεκαὶἡὑπὸδτῇὑπὸδἐστιν whole of is equal to the two (angles) and ἴση. αἱτρεῖςἄρααἱὑπὸδ,δ,ἴσαιἀλλήλαις. ut, the external (angle) is equal to εἰσίν. καὶἐπεὶἴσηἐστὶνἡὑπὸδγωνίατῇὑπὸδ, and [Prop. 1.32]. Thus, is also equal to ἴσηἐστὶκαὶπλευρὰἡδπλευρᾷτῇδ.ἀλλὰἡδτῇ. ut, is equal to, since the side is ὑπόκειταιἴση καὶἡἄρατῇδἐστινἴση ὥστεκαὶ also equal to [Prop. 1.5]. So that is also equal γωνίαἡὑπὸδγωνίᾳτῇὑπὸδἐστινἴση αἱἄρα to. Thus, the three (angles),, and ὑπὸδ,δτῆςὑπὸδεἰσιδιπλασίους. ἴσηδὲἡ are equal to one another. nd since angle is equal ὑπὸδταῖςὑπὸδ,δ καὶἡὑπὸδἄρατῆςὑπὸ to, side is also equal to side [Prop. 1.6]. Δἐστιδιπλῆ.ἴσηδὲἡὑπὸΔἑκατέρᾳτῶνὑπὸΔ, ut, was assumed (to be) equal to. Thus, Δ καὶἑκατέραἄρατῶνὑπὸδ,δτῆςὑπὸδ is also equal to. So that angle is also equal to ἐστι διπλῆ. angle [Prop. 1.5]. Thus, and is double Ισοσκελὲς ἄρα τρίγωνον συνέσταται τὸ Δ ἔχον. ut (is) equal to and. Thus, ἑκατέραντῶνπρὸςτῇδβάσειγωνιῶνδιπλασίονατῆς is also double. nd (is) equal to to λοιπῆς ὅπερ ἔδει ποιῆσαι. each of and. Thus, and are each double. Thus, the isosceles triangle has been constructed having each of the angles at the base double the remaining (angle). (Which is) the very thing it was required to do.. Proposition 11 ἰς τὸν δοθέντα κύκλον πεντάγωνον ἰσόπλευρόν τε καὶ To inscribe an equilateral and equiangular pentagon 119
ἰσογώνιονἐγγράψαι. in a given circle. Η Θ στω ὁ δοθεὶς κύκλος ὁ Δ δεῖ δὴ εἰς τὸν Let be the given circle. So it is required to Δ κύκλον πεντάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον inscribed an equilateral and equiangular pentagon in cir- ἐγγράψαι. cle. κκείσθω τρίγωνον ἰσοσκελὲς τὸ ΗΘ διπλασίονα Let the the isosceles triangle GH be set up hav- ἔχονἑκατέραντῶνπρὸςτοῖςη,θγωνιῶντῆςπρὸςτῷ, ing each of the angles at G and H double the (angle) καὶ ἐγγεγράφθω εἰς τὸν Δ κύκλον τῷ ΗΘ τριγώνῳ at [Prop. 4.10]. nd let triangle, equiangular ἰσογώνιοντρίγωνοντὸδ,ὥστετῇμὲνπρὸςτῷγωνίᾳ to GH, have been inscribed in circle, such ἴσηνεἶναιτὴνὑπὸδ,ἑκατέρανδὲτῶνπρὸςτοῖςη,θ that is equal to the angle at, and the (angles) ἴσηνἑκατέρᾳτῶνὑπὸδ,δ καὶἑκατέραἄρατῶν at G and H (are) equal to and, respectively ὑπὸ Δ, Δ τῆς ὑπὸ Δ ἐστι διπλῆ. τετμήσθω δὴ [Prop. 4.2]. Thus, and are each double ἑκατέρατῶνὑπὸδ,δδίχαὑπὸἑκατέραςτῶν,. So let and have been cut in half by Δ εὐθειῶν, καὶ ἐπεζεύχθωσαν αἱ,, Δ,. the straight-lines and, respectively [Prop. 1.9]. πεὶ οὖν ἑκατέρα τῶν ὑπὸ Δ, Δ γωνιῶν δι- nd let,, and have been joined. πλασίων ἐστὶ τῆς ὑπὸ Δ, καὶ τετμημέναι εἰσὶ δίχα ὑπὸ Therefore, since angles and are each douτῶν,δεὐθειῶν,αἱπέντεἄραγωνίαιαἱὑπὸδ, ble, and are cut in half by the straight-lines and, Δ, Δ, Δ ἴσαι ἀλλήλαις εἰσίν. αἱ δὲ ἴσαι, the five angles,,,, and γωνίαι ἐπὶ ἴσων περιφερειῶν βεβήκασιν αἱ πέντε ἄρα πε- are thus equal to one another. nd equal angles stand ριφέρειαι αἱ,, Δ, Δ, ἴσαι ἀλλήλαις εἰσίν. ὑπὸ upon equal circumferences [Prop. 3.26]. Thus, the five δὲ τὰς ἴσας περιφερείας ἴσαι εὐθεῖαι ὑποτείνουσιν αἱ πέντε circumferences,,,, and are equal to ἄρα εὐθεῖαι αἱ,, Δ, Δ, ἴσαι ἀλλήλαις εἰσίν one another [Prop. 3.29]. Thus, the pentagon ἰσόπλευρον ἄρα ἐστὶ τὸ Δ πεντάγωνον. λέγω δή, is equilateral. So I say that (it is) also equiangular. or ὅτικαὶἰσογώνιον. ἐπεὶγὰρἡπεριφέρειατῇδπε- since the circumference is equal to the circumferριφερείᾳἐστὶνἴση,κοινὴπροσκείσθωἡδ ὅληἄραἡ ence, let have been added to both. Thus, the Δ περιφέρια ὅλῃ τῇ Δ περιφερείᾳ ἐστὶν ἴση. καὶ whole circumference is equal to the whole cirβέβηκεν ἐπὶ μὲν τῆς Δ περιφερείας γωνία ἡ ὑπὸ Δ, cumference. nd the angle stands upon ἐπὶ δὲ τῆς Δ περιφερείας γωνία ἡ ὑπὸ καὶ ἡ ὑπὸ circumference, and angle upon circumfer- ἄραγωνίατῇὑπὸδἐστινἴση. διὰτὰαὐτὰδὴ ence. Thus, angle is also equal to καὶ ἑκάστη τῶν ὑπὸ, Δ, Δ γωνιῶν ἑκατέρᾳ τῶν [Prop. 3.27]. So, for the same (reasons), each of the an- ὑπὸ,δἐστινἴση ἰσογώνιονἄραἐστὶτὸδ gles,, and is also equal to each of πεντάγωνον. ἐδείχθη δὲ καὶ ἰσόπλευρον. and. Thus, pentagon is equiangular. nd ἰς ἄρα τὸν δοθέντα κύκλον πεντάγωνον ἰσόπλευρόν τε it was also shown (to be) equilateral. καὶ ἰσογώνιον ἐγγέγραπται ὅπερ ἔδει ποιῆσαι. Thus, an equilateral and equiangular pentagon has been inscribed in the given circle. (Which is) the very thing it was required to do.. Proposition 12 Περὶ τὸν δοθέντα κύκλον πεντάγωνον ἰσόπλευρόν τε To circumscribe an equilateral and equiangular penκαὶἰσογώνιονπεριγράψαι. tagon about a given circle. G H 120
Η G Θ Μ H M Κ Λ K L στω ὁ δοθεὶς κύκλος ὁ Δ δεῖ δὲ περὶ τὸν Let be the given circle. So it is required Δ κύκλον πεντάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον to circumscribe an equilateral and equiangular pentagon περιγράψαι. about circle. Νενοήσθωτοῦἐγγεγραμμένουπενταγώνουτῶνγωνιῶν Let,,,, and have been conceived as the anσημεῖατὰ,,,δ,,ὥστεἴσαςεἶναιτὰς,, gular points of a pentagon having been inscribed (in cir- Δ, Δ, περιφερείας καὶ διὰ τῶν,,, Δ, cle ) [Prop. 3.11], such that the circumferences ἤχθωσαν τοῦ κύκλου ἐφαπτόμεναι αἱ ΗΘ, ΘΚ, ΚΛ, ΛΜ,,,,, and are equal. nd let GH, HK, ΜΗ,καὶεἰλήφθωτοῦΔκύκλουκέντροντὸ,καὶ KL, LM, and MG have been drawn through (points), ἐπεζεύχθωσαναἱ,κ,,λ,δ.,,, and (respectively), touching the circle. nd ΚαὶἐπεὶἡμὲνΚΛεὐθεῖαἐφάπτεταιτοῦΔκατὰ let the center of the circle have been found τὸ,ἀπὸδὲτοῦκέντρουἐπὶτὴνκατὰτὸἐπαφὴν [Prop. 3.1]. nd let, K,, L, and have ἐπέζευκταιἡ,ἡἄρακάθετόςἐστινἐπὶτὴνκλ ὀρθὴ been joined. ἄραἐστὶνἑκατέρατῶνπρὸςτῷγωνιῶν. διὰτὰαὐτὰδὴ nd since the straight-line KL touches (circle) καὶαἱπρὸςτοῖς,δσημείοιςγωνίαιὀρθαίεἰσιν.καὶἐπεὶ at, and has been joined from the center to the ὀρθήἐστινἡὑπὸκγωνία,τὸἄραἀπὸτῆςκἴσονἐστὶ point of contact, is thus perpendicular to KL τοῖςἀπὸτῶν,κ.διὰτὰαὐτὰδὴκαὶτοῖςἀπὸτῶν, [Prop. 3.18]. Thus, each of the angles at is a right- ΚἴσονἐστὶτὸἀπὸτῆςΚ ὥστετὰἀπὸτῶν,κ angle. So, for the same (reasons), the angles at and τοῖςἀπὸτῶν,κἐστινἴσα,ὧντὸἀπὸτῆςτῷἀπὸ are also right-angles. nd since angle K is a rightτῆςἐστινἴσον λοιπὸνἄρατὸἀπὸτῆςκτῷἀπὸτῆς angle, the (square) on K is thus equal to the (sum of Κἐστινἴσον. ἴσηἄραἡκτῇκ.καὶἐπεὶἴσηἐστὶν the squares) on and K [Prop. 1.47]. So, for the ἡτῇ,καὶκοινὴἡκ,δύοδὴαἱ,κδυσὶταῖς same (reasons), the (square) on K is also equal to the,κἴσαιεἰσίν καὶβάσιςἡκβάσειτῇκ[ἐστιν]ἴση (sum of the squares) on and K. So that the (sum γωνίαἄραἡμὲνὑπὸκ[γωνίᾳ]τῇὑπὸκἐστινἴση of the squares) on and K is equal to the (sum of ἡδὲὑπὸκτῇὑπὸκ διπλῆἄραἡμὲνὑπὸτῆς the squares) on and K, of which the (square) on ὑπὸκ,ἡδὲὑπὸκτῆςὑπὸκ.διὰτὰαὐτὰδὴκαὶ is equal to the (square) on. Thus, the remain- ἡμὲνὑπὸδτῆςὑπὸλἐστιδιπλῆ,ἡδὲὑπὸδλ ing (square) on K is equal to the remaining (square) τῆςὑπὸλ.καὶἐπεὶἴσηἐστὶνἡπεριφέρειατῇδ, on K. Thus, K (is) equal to K. nd since is ἴσηἐστὶκαὶγωνίαἡὑπὸτῇὑπὸδ.καίἐστινἡ equal to, and K (is) common, the two (straightμὲνὑπὸτῆςὑπὸκδιπλῆ,ἡδὲὑπὸδτῆςὑπὸ lines), K are equal to the two (straight-lines), Λ ἴσηἄρακαὶἡὑπὸκτῇὑπὸλ ἐστὶδὲκαὶἡ K. nd the base K [is] equal to the base K. Thus, ὑπὸκγωνίατῇὑπὸλἴση. δύοδὴτρίγωνάἐστιτὰ angle K is equal to [angle] K [Prop. 1.8]. nd Κ,Λτὰςδύογωνίαςταῖςδυσὶγωνίαιςἴσαςἔχοντα K (is equal) to K [Prop. 1.8]. Thus, (is) καὶμίανπλευρὰνμιᾷπλευρᾷἴσηνκοινὴναὐτῶντὴν double K, and K (is double) K. So, for the καὶτὰςλοιπὰςἄραπλευρὰςταῖςλοιπαῖςπλευραῖςἴσαςἕξει same (reasons), is also double L, and L (is καὶ τὴν λοιπὴν γωνίαν τῇ λοιπῇ γωνίᾳ ἴση ἄρα ἡ μὲν Κ also double) L. nd since circumference is equal εὐθεῖατῇλ,ἡδὲὑπὸκγωνίατῇὑπὸλ.καὶἐπεὶἴση to, angle is also equal to [Prop. 3.27]. ἐστὶνἡκτῇλ,διπλῆἄραἡκλτῆςκ.διὰτὰαὐταδὴ nd is double K, and (is double) L. δειχθήσεταικαὶἡθκτῆςκδιπλῆ.καίἐστινἡκτῇκ Thus, K is also equal to L. nd angle K is also ἴση καὶἡθκἄρατῇκλἐστινἴση.ὁμοίωςδὴδειχθήσεται equal to L. So, K and L are two triangles hav- 121
καὶἑκάστητῶνθη,ημ,μλἑκατέρᾳτῶνθκ,κλἴση 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). Thus, ὅτι καὶ ἰσογώνιον. ἐπεὶ γὰρ ἴση ἐστὶν ἡ ὑπὸ Κ γωνία τῇ they will also have the remaining sides equal to the (cor- ὑπὸ Λ, καὶ ἐδείχθη τῆς μὲν ὑπὸ Κ διπλῆ ἡ ὑπὸ ΘΚΛ, responding) remaining sides, and the remaining angle to τῆς δὲ ὑπὸ Λ διπλῆ ἡ ὑπὸ ΚΛΜ, καὶ ἡ ὑπὸ ΘΚΛ ἄρα the remaining angle [Prop. 1.26]. Thus, the straight-line τῇὑπὸκλμἐστινἴση.ὁμοίωςδὴδειχθήσεταικαὶἑκάστη K (is) equal to L, and the angle K to L. nd τῶνὑπὸκθη,θημ,ημλἑκατέρᾳτῶνὑπὸθκλ,κλμ since K is equal to L, KL (is) thus double K. So, ἴση αἱπέντεἄραγωνίαιαἱὑπὸηθκ,θκλ,κλμ,λμη, for the same (reasons), it can be shown that HK (is) also ΜΗΘἴσαιἀλλήλαιςεἰσίν.ἰσογώνιονἄραἐστὶτὸΗΘΚΛΜ double K. nd K is equal to K. Thus, HK is also πεντάγωνον. ἐδείχθη δὲ καὶ ἰσόπλευρον, καὶ περιγέγραπται equal to KL. So, similarly, each of HG, GM, and M L περὶτὸνδκύκλον. 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. or since angle K is equal to L, and HKL was shown (to be) double K, and KLM double L, 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 are equal to one another. Thus, the pentagon GHKLM is equiangular. nd it was also shown (to be) equilateral, and has been circumscribed about circle. [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 ἰςτὸδοθὲνπεντάγωνον,ὅἐστινἰσόπλευρόντεκαὶ To inscribe a circle in a given pentagon, which is equi- ἰσογώνιον,κύκλονἐγγράψαι. lateral and equiangular. Η Μ G M Θ Λ H L Κ K στω τὸ δοθὲν πεντάγωνον ἰσόπλευρόν τε καὶ ἰσογώνι- Let be the given equilateral and equiangular οντὸδ δεῖδὴεἰςτὸδπεντάγωνονκύκλον pentagon. So it is required to inscribe a circle in pentagon ἐγγράψαι.. ΤετμήσθωγὰρἑκατέρατῶνὑπὸΔ,Δγωνιῶν or let angles and have each been cut δίχαὑπὸἑκατέραςτῶν,δεὐθειῶν καὶἀπὸτοῦ in half by each of the straight-lines and (reσημείου, καθ ὃ συμβάλλουσιν ἀλλήλαις αἱ, Δ εὐθεῖαι, spectively) [Prop. 1.9]. nd from the point, at which ἐπεζεύχθωσαναἱ,,εὐθεῖαι. καὶἐπεὶἴσηἐστὶν the straight-lines and meet one another, let the 122
ἡτῇδ,κοινὴδὲἡ,δύοδὴαἱ,δυσὶταῖς straight-lines,, and have been joined. nd Δ,ἴσαιεἰσίν καὶγωνίαἡὑπὸγωνίᾳτῇὑπὸ since is equal to, and (is) common, the two Δ[ἐστιν]ἴση βάσιςἄραἡβάσειτῇδἐστινἴση, (straight-lines), are equal to the two (straightκαὶτὸτρίγωνοντῷδτριγώνῳἐστινἴσον,καὶαἱ lines),. nd angle [is] equal to angle λοιπαὶγωνίαιταῖςλοιπαῖςγωνίαιςἴσαιἔσονται,ὑφ ἃςαἱ. Thus, the base is equal to the base, and ἴσαιπλευραὶὑποτείνουσιν ἴσηἄραἡὑπὸγωνίατῇ triangle is equal to triangle, and the remain- ὑπὸδ.καὶἐπεὶδιπλῆἐστινἡὑπὸδτῆςὑπὸδ, ing angles will be equal to the (corresponding) remain- ἴσηδὲἡμὲνὑπὸδτῇὑπὸ,ἡδὲὑπὸδτῇὑπὸ ing angles which the equal sides subtend [Prop. 1.4].,καὶἡὑπὸἄρατῆςὑπὸἐστιδιπλῆ ἴση Thus, angle (is) equal to. nd since ἄραἡὑπὸγωνίατῇὑπὸ ἡἄραὑπὸγωνία is double, and (is) equal to, and δίχατέτμηταιὑπὸτῆςεὐθείας.ὁμοίωςδὴδειχθήσεται, to, is thus also double. Thus, angle ὅτικαὶἑκατέρατῶνὑπὸ,δδίχατέτμηταιὑπὸ is equal to. Thus, angle has been cut ἑκατέραςτῶν,εὐθειῶν. ἤχθωσανδὴἀπὸτοῦ in half by the straight-line. So, similarly, it can be σημείουἐπὶτὰς,,δ,δ, εὐθείαςκάθετοι shown that and have been cut in half by the αἱη,θ,κ,λ,μ.καὶἐπεὶἴσηἐστὶνἡὑπὸθ straight-lines and, respectively. So let G, H, γωνίατῇὑπὸκ,ἐστὶδὲκαὶὀρθὴἡὑπὸθ[ὀρθῇ] K, L, and M have been drawn from point, perτῇὑπὸκἴση,δύοδὴτρίγωνάἐστιτὰθ,κτὰς pendicular to the straight-lines,,,, and δύο γωνίας δυσὶ γωνίαις ἴσας ἔχοντα καὶ μίαν πλευρὰν μιᾷ (respectively) [Prop. 1.12]. nd since angle H πλευρᾷἴσηνκοινὴναὐτῶντὴνὑποτείνουσανὑπὸμίαν is equal to K, and the right-angle H is also equal τῶνἴσωνγωνιῶν καὶτὰςλοιπὰςἄραπλευρὰςταῖςλοιπαῖς to the [right-angle] K, H and K are two triπλευραῖςἴσαςἕξει ἴσηἄραἡθκάθετοςτῂκκαθέτῳ. angles having two angles equal to two angles, and one ὁμοίωςδὴδειχθήσεται,ὅτικαὶἑκάστητῶνλ,μ,η side equal to one side, (namely) their common (side), ἑκατέρᾳτῶνθ,κἴσηἐστίν αἱπέντεἄραεὐθεῖαιαἱη, subtending one of the equal angles. Thus, they will also Θ,Κ,Λ,Μἴσαιἀλλήλαιςεἰσίν. ὁἄρακέντρῳτῷ 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 ἥξεικαὶδιὰτῶνλοιπῶνσημείωνκαὶἐφάψεταιτῶν,, H (is) equal to the perpendicular K. So, similarly, it Δ,Δ,εὐθειῶνδιὰτὸὀρθὰςεἶναιτὰςπρὸςτοῖςΗ, can be shown that L, M, and G are each equal to Θ,Κ,Λ,Μσημείοιςγωνίας. εἰγὰροὐκἐφάψεταιαὐτῶν, each of H and K. Thus, the five straight-lines G, ἀλλὰτεμεῖαὐτάς,συμβήσεταιτὴντῇδιαμέτρῳτοῦκύκλου H, K, L, and M are equal to one another. Thus, πρὸςὀρθὰςἀπ ἄκραςἀγομένηνἐντὸςπίπτειντοῦκύκλου the circle drawn with center, and radius one of G, H, ὅπερἄτοπονἐδείχθη.οὐκἄραὁκέντρῳτῷδιαστήματιδὲ K, L, or M, will also go through the remaining points, ἑνὶτῶνη,θ,κ,λ,μσημείωνγραφόμενοςκύκλοςτεμεῖ and will touch the straight-lines,,,, and τὰς,,δ,δ,εὐθείας ἐφάψεταιἄρααὐτῶν., on account of the angles at points G, H, K, L, and γεγράφθωὡςὁηθκλμ. M being right-angles. or if it does not touch them, but ἰς ἄρα τὸ δοθὲν πεντάγωνον, ὅ ἐστιν ἰσόπλευρόν τε cuts them, it follows that a (straight-line) drawn at rightκαὶ 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) absurd [Prop. 3.16]. Thus, the circle drawn with center, and radius one of G, H, K, L, or M, does not cut the straight-lines,,,, or. Thus, it will touch them. Let it have been drawn, like GHKLM (in the figure). Thus, a circle has been inscribed in the given pentagon 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,κύκλονπεριγράψαι. is equilateral and equiangular. στω τὸ δοθὲν πεντάγωνον, ὅ ἐστιν ἰσόπλευρόν τε καὶ Let be the given pentagon which is equilat- 123
ἰσογώνιον, τὸ Δ δεῖ δὴ περὶ τὸ Δ πεντάγωνον eral and equiangular. So it is required to circumscribe a κύκλονπεριγράψαι. circle about the pentagon. Τετμήσθω δὴ ἑκατέρα τῶν ὑπὸ Δ, Δ γωνιῶν So let angles and have been cut in half by δίχα ὑπὸ ἑκατέρας τῶν, Δ, καὶ ἀπὸ τοῦ σημείου, the (straight-lines) and, respectively [Prop. 1.9]. καθ ὃσυμβάλλουσιναἱεὐθεῖαι, ἐπὶτὰ,,σημεῖα nd let the straight-lines,, and have been ἐπεζεύχθωσανεὐθεῖαιαἱ,,.ὁμοίωςδὴτῷπρὸ joined from point, at which the straight-lines meet, τούτου δειχθήσεται, ὅτι καὶ ἑκάστη τῶν ὑπὸ,, to the points,, and (respectively). So, similarly, Δγωνιῶνδίχατέτμηταιὑπὸἑκάστηςτῶν,, to the (proposition) before this (one), it can be shown εὐθειῶν.καὶἐπεὶἴσηἐστὶνἡὑπὸδγωνίατῇὑπὸδ, that angles,, and have also been cut καίἐστιτῆςμὲνὑπὸδἡμίσειαἡὑπὸδ,τῆςδὲὑπὸ in half by the straight-lines,, and, respec- ΔἡμίσειαἡὑπὸΔ,καὶἡὑπὸΔἄρατῇὑπὸΔ tively. nd since angle is equal to, and ἐστινἴση ὥστεκαὶπλευρὰἡπλευρᾷτῇδἐστινἴση. is half of, and half of, is thus ὁμοίωςδὴδειχθήσεται,ὅτικαὶἑκάστητῶν,, also equal to. So that side is also equal to side ἑκατέρᾳτῶν,δἐστιν ἴση αἱπέντεἄραεὐθεῖαιαἱ [Prop. 1.6]. So, similarly, it can be shown that,,,, Δ, ἴσαι ἀλλήλαις εἰσίν. ὀ ἄρα κέντρῳ, and are also each equal to each of and. τῷκαὶδιαστήματιἑνὶτῶν,,,δ,κύκλος Thus, the five straight-lines,,,, and γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν σημείων καὶ ἔσται πε- are equal to one another. Thus, the circle drawn with ριγεγραμμένος.περιγεγράφθωκαὶἔστωὁδ. center, and radius one of,,,, or, 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, and let it be. 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 ἰς τὸν δοθέντα κύκλον ἑξάγωνον ἰσόπλευρόν τε καὶ To inscribe an equilateral and equiangular hexagon in ἰσογώνιονἐγγράψαι. a given circle. στω ὁ δοθεὶς κύκλος ὁ Δ δεῖ δὴ εἰς τὸν Let be the given circle. So it is required to Δκύκλονἑξάγωνονἰσόπλευρόντεκαὶἰσογώνιον inscribe an equilateral and equiangular hexagon in circle ἐγγράψαι.. Ηχθω τοῦ Δ κύκλου διάμετρος ἡ Δ, καὶ Let the diameter of circle have been εἰλήφθω τὸ κέντρον τοῦ κύκλουτὸ Η, καὶ κέντρῳ μὲν drawn, and let the center G of the circle have been τῷδδιαστήματιδὲτῷδηκύκλοςγεγράφθωὁηθ, found [Prop. 3.1]. nd let the circle GH have been καὶἐπιζευχθεῖσαιαἱη,ηδιήχθωσανἐπὶτὰ,σημεῖα, drawn, with center, and radius G. nd G and G καὶἐπεζεύχθωσαναἱ,,δ,δ,, λέγω,ὅτι being joined, let them have been drawn across (the cir- 124
τὸδἑξάγωνονἰσόπλευρόντέἐστικαὶἰσογώνιον. cle) to points and (respectively). nd let,,,,, and have been joined. I say that the hexagon is equilateral and equiangular. Θ H Η G πεὶ γὰρ τὸ Η σημεῖον κέντρον ἐστὶ τοῦ Δ or since point G is the center of circle, κύκλου,ἴσηἐστὶνἡητῇηδ.πάλιν,ἐπεὶτὸδσημεῖον G is equal to G. gain, since point is the cenκέντρονἐστὶτοῦηθκύκλου,ἴσηἐστὶνἡδτῇδη. ter of circle GH, is equal to G. ut, G was ἀλλ ἡητῇηδἐδείχθηἴση καὶἡηἄρατῇδἴση shown (to be) equal to G. Thus, G is also equal to ἐστίν ἰσόπλευρον ἄρα ἐστὶ τὸ ΗΔ τρίγωνον καὶ αἱ τρεῖς. Thus, triangle G is equilateral. Thus, its three ἄρααὐτοῦγωνίαιαἱὑπὸηδ,ηδ,δηἴσαιἀλλήλαις angles G, G, and G are also equal to one anεἰσίν, ἐπειδήπερ τῶν ἰσοσκελῶν τριγώνων αἱ πρὸς τῇ βάσει other, inasmuch as the angles at the base of isosceles triγωνίαι ἴσαι ἀλλήλαις εἰσίν καί εἰσιν αἱ τρεῖς τοῦ τριγώνου angles are equal to one another [Prop. 1.5]. nd the γωνίαι δυσὶν ὀρθαῖς ἴσαι ἡ ἄρα ὑπὸ ΗΔ γωνία τρίτον ἐστὶ three angles of the triangle are equal to two right-angles δύο ὀρθῶν. ὁμοίως δὴ δειχθήσεται καὶ ἡ ὑπὸ ΔΗ τρίτον [Prop. 1.32]. Thus, angle G is one third of two rightδύοὀρθῶν.καὶἐπεὶἡηεὐθεῖαἐπὶτὴνσταθεῖσατὰς angles. So, similarly, G can also be shown (to be) ἐφεξῆς γωνίας τὰς ὑπὸ Η, Η δυσὶν ὀρθαῖς ἴσας ποιεῖ, one third of two right-angles. nd since the straight-line καὶλοιπὴἄραἡὑπὸητρίτονἐστὶδύοὀρθῶν αἱἄρα G, standing on, makes adjacent angles G and ὑπὸηδ,δη,ηγωνίαιἴσαιἀλλήλαιςεἰσίν ὥστεκαὶ G equal to two right-angles [Prop. 1.13], the remainαἱ κατὰ κορυφὴν αὐταῖς αἱ ὑπὸ Η, Η, Η ἴσαι εἰσὶν ing angle G is thus also one third of two right-angles. [ταῖςὑπὸηδ,δη,η].αἱἓξἄραγωνίαιαἱὑπὸηδ, Thus, angles G, G, and G are equal to one an- ΔΗ, Η, Η, Η, Η ἴσαι ἀλλήλαις εἰσίν. αἱ δὲ other. nd hence the (angles) opposite to them G, ἴσαιγωνίαιἐπὶἴσωνπεριφερειῶνβεβήκασιν αἱἓξἄραπε- G, and G are also equal [to G, G, and ριφέρειαι αἱ,, Δ, Δ,, ἴσαι ἀλλήλαις εἰσίν. G (respectively)] [Prop. 1.15]. Thus, the six angles ὑπὸδὲτὰςἴσαςπεριφερείαςαἱἴσαιεὐθεῖαιὑποτείνουσιν G, G, G, G, G, and G are equal αἱ ἓξ ἄρα εὐθεῖαι ἴσαι ἀλλήλαις εἰσίν ἰσόπλευρον ἄρα ἐστὶ to one another. nd equal angles stand on equal cirτο Δ ἑξάγωνον. λέγω δή, ὅτι καὶ ἰσογώνιον. ἐπεὶ cumferences [Prop. 3.26]. Thus, the six circumferences γὰρἴσηἐστὶνἡπεριφέρειατῇδπεριφερείᾳ, κοινὴ,,,,, and are equal to one anπροσκείσθω ἡ Δ περιφέρεια ὅλη ἄρα ἡ Δ ὅλῃ other. nd equal circumferences are subtended by equal τῇ Δ ἐστιν ἴση καὶ βέβηκεν ἐπὶ μὲν τῆς Δ straight-lines [Prop. 3.29]. Thus, the six straight-lines περιφερείαςἡὑπὸδγωνία,ἐπὶδὲτῆςδπερι- (,,,,, and ) are equal to one φερείας ἡ ὑπὸ γωνία ἴση ἄρα ἡ ὑπὸ γωνία τῇ another. Thus, hexagon is equilateral. So, ὑπὸ Δ. ὁμοίως δὴ δειχθήσεται, ὅτι καὶ αἱ λοιπαὶ γωνίαι I say that (it is) also equiangular. or since circumferτοῦ Δ ἑξαγώνου κατὰ μίαν ἴσαι εἰσὶν ἑκατέρᾳ τῶν ence is equal to circumference, let circumference ὑπὸ,δγωνιῶν ἰσογώνιονἄραἐστὶτὸδ have been added to both. Thus, the whole of ἑξάγωνον. ἐδείχθη δὲ καὶ ἰσόπλευρον καὶ ἐγγέγραπται εἰς is equal to the whole of. nd angle τὸνδκύκλον. stands on circumference, and angle ἰς ἄρα τὸν δοθέντα κύκλον ἑξάγωνον ἰσόπλευρόν τε on circumference. Thus, angle is equal 125
καὶ ἰσογώνιον ἐγγέγραπται ὅπερ ἔδει ποιῆσαι. to [Prop. 3.27]. Similarly, it can also be shown that the remaining angles of hexagon are individually equal to each of the angles and. Thus, hexagon is equiangular. nd it was also shown (to be) equilateral. nd it has been inscribed in circle. Thus, an equilateral and equiangular hexagon has been inscribed in the given circle. (Which is) the very thing it was required to do. È Ö Ñ. orollary κ δὴ τούτου φανερόν, ὅτι ἡ τοῦ ἑξαγώνου πλευρὰ ἴση So, from this, (it is) manifest that a side of the ἐστὶτῇἐκτοῦκέντρουτοῦκύκλου. hexagon is equal to the radius of the circle. Ομοίως δὲ τοῖς ἐπὶ τοῦ πενταγώνου ἐὰν διὰ τῶν κατὰ nd similarly to a pentagon, if we draw tangents τὸν κύκλον διαιρέσεων ἐφαπτομένας τοῦ κύκλου ἀγάγωμεν, to the circle through the (sixfold) divisions of the (cirπεριγραφήσεται περὶ τὸν κύκλον ἑξάγωνον ἰσόπλευρόν cumference of the) circle, an equilateral and equianguτε καὶ ἰσογώνιον ἀκολούθως τοῖς ἐπὶ τοῦ πενταγώνου lar hexagon can be circumscribed about the circle, analoεἰρημένοις. καὶ ἔτι διὰ τῶν ὁμοίων τοῖς ἐπὶ τοῦ πενταγώνου gously to the aforementioned pentagon. nd, 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 hexagon. (Which is) the very thing it was required to do. See the footnote to Prop. 4.6.. Proposition 16 ἰς τὸν δοθέντα κύκλον πεντεκαιδεκάγωνον ἰσόπλευρόν To inscribe an equilateral and equiangular fifteenτεκαὶἰσογώνιονἐγγράψαι. sided figure in a given circle. στω ὁ δοθεὶς κύκλος ὁ Δ δεῖ δὴ εἰς τὸν Δ Let be the given circle. So it is required to inκύκλον πεντεκαιδεκάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον scribe an equilateral and equiangular fifteen-sided figure ἐγγράψαι. in circle. γγεγράφθω εἰς τὸν Δ κύκλον τριγώνου μὲν ἰσο- Let the side of an equilateral triangle inscribed πλεύρουτοῦεἰςαὐτὸνἐγγραφομένουπλευρὰἡ,πεν- in (the circle) [Prop. 4.2], and (the side) of an (inταγώνου δὲ ἰσοπλεύρου ἡ οἵων ἄρα ἐστὶν ὁ Δ scribed) equilateral pentagon [Prop. 4.11], have been inκύκλος ἴσων τμήματων δεκαπέντε, τοιούτων ἡ μὲν scribed in circle. Thus, just as the circle περιφέρεια τρίτον οὖσα τοῦ κύκλου ἔσται πέντε, ἡ δὲ is (made up) of fifteen equal pieces, the circumference περιφέρειαπέμτονοὖσατοῦκύκλουἔσταιτριῶν λοιπὴἄρα, being a third of the circle, will be (made up) of five 126
ἡ τῶν ἴσων δύο. τετμήσθω ἡ δίχα κατὰ τὸ such (pieces), and the circumference, being a fifth of ἑκατέρα ἄρα τῶν, περιφερειῶν πεντεκαιδέκατόν ἐστι the circle, will be (made up) of three. Thus, the remainτοῦ Δ κύκλου. der (will be made up) of two equal (pieces). Let (cir- ὰν ἄρα ἐπιζεύξαντες τὰς, ἴσας αὐταῖς κατὰ τὸ cumference) have been cut in half at [Prop. 3.30]. συνεχὲς εὐθείας ἐναρμόσωμεν εἰς τὸν Δ[] κύκλον, Thus, each of the circumferences and is one fif- ἔσται εἰς αὐτὸν ἐγγεγραμμένον πεντεκαιδεκάγωνον ἰσόπλευ- teenth of the circle. ρόν τε καὶ ἰσογώνιον ὅπερ ἔδει ποιῆσαι. Thus, if, joining and, we continuously in- Ομοίως δὲ τοῖς ἐπὶ τοῦ πενταγώνου ἐὰν διὰ τῶν sert straight-lines equal to them into circle [] 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 nd similarly to the pentagon, if we draw tangents to δοθὲν πεντεκαιδεκάγωνον κύκλον ἐγγράψομέν τε καὶ πε- the circle through the (fifteenfold) divisions of the (cirριγράψομεν ὅπερ ἔδει ποιῆσαι. cumference of the) circle, we can circumscribe an equilateral and equiangular fifteen-sided figure about the circle. nd, 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