Definitions. . Proposition 1

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1 ÎÇÖÓ. efinitions αʹ. Ισοι κύκλοι εἰσίν, ὧν αἱ διάμετροι ἴσαι εἰσίν, ἢ ὧν αἱ 1. qual circles are (circles) whose diameters are ἐκ τῶν κέντρων ἴσαι εἰσίν. equal, or whose (distances) from the centers (to the cirβʹ.ὐθεῖακύκλουἐφάπτεσθαιλέγεται,ἥτιςἁπτομένη cumferences) are equal (i.e., whose radii are equal). τοῦ κύκλου καὶ ἐκβαλλομένη οὐ τέμνει τὸν κύκλον. 2. straight-line said to touch a circle is any (straightγʹ.κύκλοιἐφάπτεσθαιἀλλήλωνλέγονταιοἵτινεςἁπτό- line) which, meeting the circle and being produced, does μενοι ἀλλήλων οὐ τέμνουσιν ἀλλήλους. not cut the circle. δʹ. ν κύκλῳ ἴσον ἀπέχειν ἀπὸ τοῦ κέντρου εὐθεῖαι 3. ircles said to touch one another are any (circles) λέγονται, ὅταν αἱ ἀπὸ τοῦ κέντρου ἐπ αὐτὰς κάθετοι which, meeting one another, do not cut one another. ἀγόμεναι ἴσαι ὦσιν. 4. In a circle, straight-lines are said to be equally far εʹ. Μεῖζον δὲ ἀπέχειν λέγεται, ἐφ ἣν ἡ μείζων κάθετος from the center when the perpendiculars drawn to them πίπτει. from the center are equal. ϛʹ. Τμῆμα κύκλου ἐστὶ τὸ περιεχόμενον σχῆμα ὑπό τε 5. nd (that straight-line) is said to be further (from εὐθείας καὶ κύκλου περιφερείας. the center) on which the greater perpendicular falls ζʹ. Τμήματος δὲ γωνία ἐστὶν ἡ περιεχομένη ὑπό τε (from the center). εὐθείας καὶ κύκλου περιφερείας. 6. segment of a circle is the figure contained by a ηʹ. ν τμήματι δὲ γωνία ἐστίν, ὅταν ἐπὶ τῆς περι- straight-line and a circumference of a circle. φερείαςτοῦτμήματοςληφθῇτισημεῖονκαὶἀπ αὐτοῦἐπὶ 7. nd the angle of a segment is that contained by a τὰ πέρατα τῆς εὐθείας, ἥ ἐστι βάσις τοῦ τμήματος, ἐπι- straight-line and a circumference of a circle. ζευχθῶσιν εὐθεῖαι, ἡ περιεχομένη γωνία ὑπὸ τῶν ἐπιζευ- 8. nd the angle in a segment is the angle contained χθεισῶν εὐθειῶν. by the joined straight-lines, when any point is taken on θʹ. Οταν δὲ αἱ περιέχουσαι τὴν γωνίαν εὐθεῖαι ἀπο- the circumference of a segment, and straight-lines are λαμβάνωσί τινα περιφέρειαν, ἐπ ἐκείνης λέγεται βεβηκέναι joined from it to the ends of the straight-line which is ἡγωνία. the base of the segment. ιʹ. Τομεὺς δὲ κύκλου ἐστίν, ὅταν πρὸς τῷ κέντρῷ τοῦ 9. nd when the straight-lines containing an angle κύκλου συσταθῇ γωνία, τὸ περιεχόμενον σχῆμα ὑπό τε τῶν cut off some circumference, the angle is said to stand τὴν γωνίαν περιεχουσῶν εὐθειῶν καὶ τῆς ἀπολαμβανομένης upon that (circumference). ὑπ αὐτῶνπεριφερείας. 10. nd a sector of a circle is the figure contained by ιαʹ. Ομοία τμήματα κύκλων ἐστὶ τὰ δεχόμενα γωνίας the straight-lines surrounding an angle, and the circum- ἴσας, ἤ ἐν οἷς αἱ γωνίαι ἴσαι ἀλλήλαις εἰσίν. ference cut off by them, when the angle is constructed at the center of a circle. 11. Similar segments of circles are those accepting equal angles, or in which the angles are equal to one another.. Proposition 1 Τοῦδοθέντοςκύκλουτὸκέντρονεὑρεῖν. To find the center of a given circle. στωὁδοθεὶςκύκλοςὁ δεῖδὴτοῦκύκλου Let be the given circle. So it is required to find τὸκέντρονεὑρεῖν. the center of circle. Διήχθω τις εἰς αὐτόν, ὡς ἔτυχεν, εὐθεῖα ἡ, καὶ Let some straight-line have been drawn through τετμήσθωδίχακατὰτὸδσημεῖον,καὶἀπὸτοῦδτῇ (), at random, and let () have been cut in half at πρὸςὀρθὰςἤχθωἡδκαὶδιήχθωἐπὶτὸ,καὶτετμήσθω point [Prop. 1.9]. nd let have been drawn from ἡδίχακατὰτὸ λέγω,ὅτιτὸκέντρονἐστὶτοῦ, at right-angles to [Prop. 1.11]. nd let () have [κύκλου]. been drawn through to. nd let have been cut in Μὴ γάρ, ἀλλ εἰ δυνατόν, ἔστω τὸ Η, καὶ ἐπεζεύχθωσαν half at [Prop. 1.9]. I say that (point) is the center of αἱη,ηδ,η.καὶἐπεὶἴσηἐστὶνἡδτῇδ,κοινὴδὲἡ the [circle]. ΔΗ,δύοδὴαἱΔ,ΔΗδύοταῖςΗΔ,Δἴσαιεἰσὶνἑκατέρα or (if) not then, if possible, let G (be the center of the ἑκατέρᾳ καὶβάσιςἡηβάσειτῇηἐστινἴση ἐκκέντρου circle), and let G, G, and G have been joined. nd γάρ γωνίαἄραἡὑπὸδηγωνίᾳτῇὑπὸηδἴσηἐστίν. since is equal to, and G (is) common, the two 70

2 ὅταν δὲ εὐθεῖα ἐπ εὐθεῖαν σταθεῖσα τὰς ἐφεξῆς γωνίας (straight-lines), G are equal to the two (straight- ἴσαςἀλλήλαιςποιῇ,ὀρθὴἑκατέρατῶνἴσωνγωνιῶνἐστιν lines), G, respectively. nd the base G is equal ὀρθὴἄραἐστὶνἡὑπὸηδ.ἐστὶδὲκαὶἡὑπὸδὀρθή to the base G. or (they are both) radii. Thus, angle ἴσηἄραἡὑπὸδτῇὑπὸηδ,ἡμείζωντῇἐλάττονι G is equal to angle G [Prop. 1.8]. nd when a ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τὸ Η κέντρον ἐστὶ τοῦ straight-line stood upon (another) straight-line make adκύκλου.ὁμοίωςδὴδείξομεν,ὅτιοὐδ ἄλλοτιπλὴντοῦ. jacent angles (which are) equal to one another, each of the equal angles is a right-angle [ef. 1.10]. Thus, G is a right-angle. nd is also a right-angle. Thus, (is) equal to G, the greater to the lesser. The very thing is impossible. Thus, (point) G is not the center of the circle. So, similarly, we can show that neither is any other (point) except. Η G Τὸ ἄρα σημεῖον κέντρον ἐστὶ τοῦ [κύκλου]. Thus, point is the center of the [circle]. È Ö Ñ. orollary κδὴτούτουφανερόν, ὅτιἐὰνἐνκύκλῳεὐθεῖάτις So, from this, (it is) manifest that if any straight-line εὐθεῖάν τινα δίχα καὶ πρὸς ὀρθὰς τέμνῃ, ἐπὶ τῆς τεμνούσης in a circle cuts any (other) straight-line in half, and at ἐστὶ τὸ κέντρον τοῦ κύκλου. ὅπερ ἔδει ποιῆσαι. right-angles, then the center of the circle is on the former (straight-line). (Which is) the very thing it was required to do. The Greek text has G,, which is obviously a mistake.. Proposition 2 ὰν κύκλου ἐπὶ τῆς περιφερείας ληφθῇ δύο τυχόντα If two points are taken at random on the circumferσημεῖα, ἡ ἐπὶ τὰ σημεῖα ἐπιζευγνυμένη εὐθεῖα ἐντὸς πεσεῖται ence of a circle then the straight-line joining the points τοῦκύκλου. will fall inside the circle. στω κύκλος ὁ, καὶ ἐπὶ τῆς περιφερείας αὐτοῦ Let be a circle, and let two points and have εἰλήφθωδύοτυχόντασημεῖατὰ, λέγω, ὅτιἡἀπὸ been taken at random on its circumference. I say that the τοῦ ἐπὶ τὸ ἐπιζευγνυμένη εὐθεῖα ἐντὸς πεσεῖται τοῦ straight-line joining to will fall inside the circle. κύκλου. or (if) not then, if possible, let it fall outside (the Μὴγάρ,ἀλλ εἰδυνατόν,πιπτέτωἐκτὸςὡςἡ,καὶ circle), like (in the figure). nd let the center of εἰλήφθωτὸκέντροντοῦκύκλου,καὶἔστωτὸδ,καὶ the circle have been found [Prop. 3.1], and let it be ἐπεζεύχθωσαναἱδ,δ,καὶδιήχθωἡδ. (at point). nd let and have been joined, and πεὶοὖνἴσηἐστὶνἡδτῇδ,ἴσηἄρακαὶγωνίαἡ let have been drawn through. ὑπὸδτῇὑπὸδ καὶἐπεὶτριγώνουτοῦδμία Therefore, since is equal to, the angle 71

3 angles to [Prop. 1.11]. I say that the center of the circle is on. Μὴγάρ,ἀλλ εἰδυνατόν,ἔστωτὸ,καὶἐπεζεύχθωἡ or (if) not, if possible, let be (the center of the. circle), and let have been joined. πεὶ[οὖν] κύκλου τοῦ ἐφάπτεταί τις εὐθεῖα ἡ Δ, [Therefore], since some straight-line touches the ἀπὸδὲτοῦκέντρουἐπὶτὴνἁφὴνἐπέζευκταιἡ,ἡἄρα circle, and has been joined from the center to κάθετόςἐστινἐπὶτὴνδ ὀρθὴἄραἐστὶνἡὑπὸ.ἐστὶ the point of contact, is thus perpendicular to δὲκαὶἡὑπὸὀρθή ἴσηἄραἐστὶνἡὑπὸτῇὑπὸ [Prop. 3.18]. Thus, is a right-angle. nd ἡἐλάττωντῇμείζονι ὅπερἐστὶνἀδύνατον. οὐκἄρα is also a right-angle. Thus, is equal to, the τὸ κέντρον ἐστὶ τοῦ κύκλου. ὁμοίως δὴ δείξομεν, lesser to the greater. The very thing is impossible. Thus, ὅτιοὐδ ἄλλοτιπλὴνἐπὶτῆς. is not the center of circle. So, similarly, we can ὰν ἄρα κύκλου ἐφάπτηταί τις εὐθεῖα, ἀπὸ δὲ τῆς ἁφῆς show that neither is any (point) other (than one) on. τῇ ἐφαπτομένῃ πρὸς ὀρθὰς εὐθεῖα γραμμὴ ἀχθῇ, ἐπὶ τῆς ἀχθείσης ἔσται τὸ κέντρον τοῦ κύκλου ὅπερ ἔδει δεῖξαι. Thus, if some straight-line touches a circle, and a straightline is drawn from the point of contact, at right-angles to the tangent, then the center (of the circle) will be on the (straight-line) so drawn. (Which is) the very thing it was required to show.. Proposition 20 νκύκλῳἡπρὸςτῷκέντρῳγωνίαδιπλασίωνἐστὶτῆς In a circle, the angle at the center is double that at the πρὸς τῇ περιφερείᾳ, ὅταν τὴν αὐτὴν περιφέρειαν βάσιν ἔχω- circumference, when the angles have the same circumferσιν αἱ γωνίαι. ence base. στωκύκλοςὁ,καὶπρὸςμὲντῷκέντρῳαὐτοῦ Let be a circle, and let be an angle at its γωνίαἔστωἡὑπὸ,πρὸςδὲτῇπεριφερείᾳἡὑπὸ, center, and (one) at (its) circumference. nd let ἐχέτωσαν δὲ τὴν αὐτὴν περιφέρειαν βάσιν τὴν λέγω, them have the same circumference base. I say that ὅτι διπλασίων ἐστὶν ἡ ὑπὸ γωνία τῆς ὑπὸ. angle is double (angle). πιζευχθεῖσαγὰρἡδιήχθωἐπὶτὸ. or being joined, let have been drawn through to πεὶοὖνἴσηἐστὶνἡτῇ,ἴσηκαὶγωνίαἡὑπὸ. τῇὑπὸ αἱἄραὑπὸ,γωνίαιτῆςὑπὸ Therefore, since is equal to, angle (is) διπλασίουςεἰσίν. ἴσηδὲἡὑπὸταῖςὑπὸ, also equal to [Prop. 1.5]. Thus, angle and καὶἡὑπὸἄρατῆςὑπὸἐστιδιπλῆ.διὰτὰ is double (angle). nd (is) equal to αὐτὰδὴκαὶἡὑπὸτῆςὑπὸἐστιδιπλῆ. ὅληἄρα and [Prop. 1.32]. Thus, is also double ἡὑπὸὅληςτῆςὑπὸἐστιδιπλῆ.. So, for the same (reasons), is also double. Thus, the whole (angle) is double the whole (angle). 90

4 Η Κεκλάσθω δὴ πάλιν, καὶ ἔστω ἑτέρα γωνία ἡ ὑπὸ Δ, So let another (straight-line) have been inflected, and καὶἐπιζευχθεῖσαἡδἐκβεβλήσθωἐπὶτὸη.ὁμοίωςδὴ let there be another angle,. nd being joined, δείξομεν,ὅτιδιπλῆἐστινἡὑπὸηγωνίατῆςὑπὸδ, let it have been produced to G. So, similarly, we can show ὧνἡὑπὸηδιπλῆἐστιτῆςὑπὸδ λοιπὴἄραἡὑπὸ that angle G is double, of which G is double διπλῆ ἐστι τῆς ὑπὸ Δ.. Thus, the remaining (angle) is double the ν κύκλῳ ἄρα ἡ πρὸς τῷ κέντρῳ γωνία διπλασίων ἐστὶ (remaining angle). τῆς πρὸς τῇ περιφερείᾳ, ὅταν τὴν αὐτὴν περιφέρειαν βάσιν Thus, in a circle, the angle at the center is double that ἔχωσιν[αἱ γωνίαι] ὅπερ ἔδει δεῖξαι. at the circumference, when [the angles] have the same circumference base. (Which is) the very thing it was required to show.. Proposition 21 νκύκλῳαἱἐντῷαὐτῷτμήματιγωνίαιἴσαιἀλλήλαις In a circle, angles in the same segment are equal to εἰσίν. one another. G στωκύκλοςὁδ,καὶἐντῷαὐτῷτμήματιτῷ Let be a circle, and let and be ΔγωνίαιἔστωσαναἱὑπὸΔ,Δ λέγω,ὅτιαἱ angles in the same segment. I say that angles ὑπὸ Δ, Δ γωνίαι ἴσαι ἀλλήλαις εἰσίν. and are equal to one another. ἰλήφθωγὰρτοῦδκύκλουτὸκέντρον,καὶἔστω or let the center of circle have been found τὸ,καὶἐπεζεύχθωσαναἱ,δ. [Prop. 3.1], and let it be (at point). nd let and ΚαὶἐπεὶἡμὲνὑπὸΔγωνίαπρὸςτῷκέντρῳἐστίν,ἡ have been joined. δὲὑπὸδπρὸςτῇπεριφερείᾳ,καὶἔχουσιτὴναὐτὴνπε- nd since angle is at the center, and at ριφέρειαν βάσιν τὴν Δ, ἡ ἄρα ὑπὸ Δ γωνία διπλασίων the circumference, and they have the same circumference ἐστὶτῆςὑπὸδ.διὰτὰαὐτὰδὴἡὑπὸδκαὶτῆςὑπὸ base, angle is thus double [Prop. 3.20]. 91

5 Δἐστιδιπλσίων ἴσηἄραἡὑπὸδτῇὑπὸδ. So, for the same (reasons), is also double. ν κύκλῳ ἄρα αἱ ἐν τῷ αὐτῷ τμήματι γωνίαι ἴσαι Thus, (is) equal to. ἀλλήλαις εἰσίν ὅπερ ἔδει δεῖξαι. Thus, in a circle, angles in the same segment are equal to one another. (Which is) the very thing it was required to show.. Proposition 22 Τῶν ἐν τοῖς κύκλοις τετραπλεύρων αἱ ἀπεναντίον γωνίαι or quadrilaterals within circles, the (sum of the) opδυσὶνὀρθαῖςἴσαιεἰσίν. posite angles is equal to two right-angles. στωκύκλοςὁδ,καὶἐναὐτῷτετράπλευρονἔστω Let be a circle, and let be a quadrilatτὸ Δ λέγω, ὅτι αἱ ἀπεναντίον γωνίαι δυσὶν ὀρθαῖς ἴσαι eral within it. I say that the (sum of the) opposite angles εἰσίν. is equal to two right-angles. πεζεύχθωσαν αἱ, Δ. Let and have been joined. πεὶ οὖν παντὸς τριγώνου αἱ τρεῖς γωνίαι δυσὶν ὀρθαῖς Therefore, since the three angles of any triangle are ἴσαι εἰσίν, τοῦ ἄρα τριγώνου αἱ τρεῖς γωνίαι αἱ ὑπὸ equal to two right-angles [Prop. 1.32], the three angles,,δυσὶνὀρθαῖςἴσαιεἰσίν.ἴσηδὲἡμὲνὑπὸ,, and of triangle are thus equal τῇὑπὸδ ἐνγὰρτῷαὐτῷτμήματίεἰσιτῷδ to two right-angles. nd (is) equal to. or ἡδὲὑπὸτῇὑπὸδ ἐνγὰρτῷαὐτῷτμήματίεἰσι they are in the same segment [Prop. 3.21]. nd τῷδ ὅληἄραἡὑπὸδταῖςὑπὸ,ἴση (is equal) to. or they are in the same seg- ἐστίν. κοινὴπροσκείσθωἡὑπὸ αἱἄραὑπὸ, ment [Prop. 3.21]. Thus, the whole of is,ταῖςὑπὸ,δἴσαιεἰσίν. ἀλλ αἱὑπὸ equal to and. Let have been added to,,δυσὶνὀρθαῖςἴσαιεἰσίν.καὶαἱὑπὸ, both. Thus,,, and are equal to Δἄραδυσὶνὀρθαῖςἴσαιεἰσίν. ὁμοίωςδὴδείξομεν,ὅτι and. ut,,, and are equal to two καὶ αἱ ὑπὸ Δ, Δ γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν. right-angles. Thus, and are also equal to two Τῶν ἄρα ἐν τοῖς κύκλοις τετραπλεύρων αἱ ἀπεναντίον right-angles. Similarly, we can show that angles γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν ὅπερ ἔδει δεῖξαι. and are also equal to two right-angles. Thus, for quadrilaterals within circles, the (sum of the) opposite angles is equal to two right-angles. (Which is) the very thing it was required to show.. Proposition 23 πὶ τῆς αὐτῆς εὐθείας δύο τμήματα κύκλων ὅμοια καὶ Two similar and unequal segments of circles cannot be ἄνισαοὐσυσταθήσεταιἐπὶτὰαὐτὰμέρη. constructed on the same side of the same straight-line. ἰ γὰρ δυνατόν, ἐπὶ τῆς αὐτῆςεὐθείαςτῆς δύο or, if possible, let the two similar and unequal segτμήματα κύκλων ὅμοια καὶ ἄνισα συνεστάτω ἐπὶ τὰ αὐτὰ ments of circles, and, have been constructed μέρητὰ,δ,καὶδιήχθωἡδ,καὶἐπεζεύχθωσαν on the same side of the same straight-line. nd let 92