LMNTS OOK 3 undamentals of Plane eometry Involving ircles 69
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, καὶ ἐπεζεύχθωσαν half at [Prop. 1.9]. I say that (point) is the center of αἱ,δ,.καὶἐπεὶἴσηἐστὶνἡδτῇδ,κοινὴδὲἡ the [circle]. Δ,δύοδὴαἱΔ,ΔδύοταῖςΔ,Δἴσαιεἰσὶνἑκατέρα or (if) not then, if possible, let (be the center of the ἑκατέρᾳ καὶβάσιςἡβάσειτῇἐστινἴση ἐκκέντρου circle), and let,, and have been joined. nd γάρ γωνίαἄραἡὑπὸδγωνίᾳτῇὑπὸδἴσηἐστίν. since is equal to, and (is) common, the two 70
ὅταν δὲ εὐθεῖα ἐπ εὐθεῖαν σταθεῖσα τὰς ἐφεξῆς γωνίας (straight-lines), are equal to the two (straight- ἴσαςἀλλήλαιςποιῇ,ὀρθὴἑκατέρατῶνἴσωνγωνιῶνἐστιν lines),, respectively. nd the base is equal ὀρθὴἄραἐστὶνἡὑπὸδ.ἐστὶδὲκαὶἡὑπὸδὀρθή to the base. or (they are both) radii. Thus, angle ἴσηἄραἡὑπὸδτῇὑπὸδ,ἡμείζωντῇἐλάττονι is equal to angle [Prop. 1.8]. nd when a ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τὸ 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, is a right-angle. nd is also a right-angle. Thus, (is) equal to, the greater to the lesser. The very thing is impossible. Thus, (point) is not the center of the circle. So, similarly, we can show that neither is any other (point) except. 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 reek text has,, 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
πλευρὰ προσεκβέβληται ἡ, μείζων ἄρα ἡ ὑπὸ Δ (is) thus also equal to [Prop. 1.5]. nd since in triγωνίατῆςὑπὸδ. ἴσηδὲἡὑπὸδ τῇὑπὸδ angle the one side,, has been produced, anμείζωνἄραἡὑπὸδτῆςὑπὸδ.ὑπὸδὲτὴνμείζονα gle (is) thus greater than [Prop. 1.16]. nd γωνίανἡμείζωνπλευρὰὑποτείνει μείζωνἄραἡδτῆς (is) equal to [Prop. 1.5]. Thus, (is) Δ. ἴσηδὲἡδτῇδ.μείζων ἄραἡδτῆςδἡ greater than. nd the greater angle is subtended ἐλάττων τῆς μείζονος ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἡ by the greater side [Prop. 1.19]. Thus, (is) greater ἀπὸτοῦἐπὶτὸἐπιζευγνυμένηεὐθεῖαἐκτὸςπεσεῖται than. nd (is) equal to. Thus, (is) τοῦ κύκλου. ὁμοίως δὴ δείξομεν, ὅτι οὐδὲ ἐπ αὐτῆς τῆς greater than, the lesser than the greater. The very περιφερείας ἐντὸς ἄρα. thing is impossible. Thus, the straight-line joining to will not fall outside the circle. So, similarly, we can show that neither (will it fall) on the circumference itself. Thus, (it will fall) inside (the circle). ὰνἄρακύκλουἐπὶτῆςπεριφερείαςληφθῇδύοτυχόντα Thus, if two points are taken at random on the cirσημεῖα, ἡ ἐπὶ τὰ σημεῖα ἐπιζευγνυμένη εὐθεῖα ἐντὸς πεσεῖται cumference of a circle then the straight-line joining the τοῦ κύκλου ὅπερ ἔδει δεῖξαι. points will fall inside the circle. (Which is) the very thing it was required to show.. Proposition 3 ὰν ἐν κύκλῳ εὐθεῖά τις διὰ τοῦ κέντρου εὐθεῖάν τινα In a circle, if any straight-line through the center cuts μὴ διὰ τοῦ κέντρου δίχα τέμνῃ, καὶ πρὸς ὀρθὰς αὐτὴν τέμνει in half any straight-line not through the center then it καὶἐὰνπρὸςὀρθὰςαὐτὴντέμνῃ,καὶδίχααὐτὴντέμνει. also cuts it at right-angles. nd (conversely) if it cuts it στω κύκλος ὁ, καὶ ἐν αὐτῷ εὐθεῖά τις διὰ τοῦ at right-angles then it also cuts it in half. κέντρου ἡ Δ εὐθεῖάν τινα μὴ διὰ τοῦ κέντρου τὴν δίχα Let be a circle, and, within it, let some straightτεμνέτω κατὰ τὸ σημεῖον λέγω, ὅτι καὶ πρὸς ὀρθὰς αὐτὴν line through the center,, cut in half some straight-line τέμνει. not through the center,, at the point. I say that ἰλήφθω γὰρ τὸ κέντρον τοῦ κύκλου, καὶ ἔστω () also cuts () at right-angles. τὸ,καὶἐπεζεύχθωσαναἱ,. or let the center of the circle have been found Καὶἐπεὶἴσηἐστὶνἡτῇ,κοινὴδὲἡ,δύοδυσὶν [Prop. 3.1], and let it be (at point), and let and ἴσαι[εἰσίν] καὶ βάσις ἡ βάσει τῇ ἴση γωνία ἄρα ἡ have been joined. ὑπὸγωνίᾳτῇὑπὸἴσηἐστίν.ὅτανδὲεὐθεῖαἐπ nd since is equal to, and (is) common, εὐθεῖαν σταθεῖσα τὰς ἐφεξῆς γωνίας ἴσας ἀλλήλαις ποιῇ, two (sides of triangle ) [are] equal to two (sides of ὀρθὴἑκατέρατῶν ἴσωνγωνιῶνἐστιν ἑκατέραἄρατῶν triangle ). nd the base (is) equal to the base ὑπὸ,ὀρθήἐστιν. ἡδἄραδιὰτοῦκέντρου. Thus, angle is equal to angle [Prop. 1.8]. οὖσα τὴν μὴ διὰ τοῦ κέντρου οὖσαν δίχα τέμνουσα καὶ nd when a straight-line stood upon (another) straightπρὸς ὀρθὰς τέμνει. line makes adjacent angles (which are) equal to one another, each of the equal angles is a right-angle [ef. 1.10]. Thus, and are each right-angles. Thus, the 72
(straight-line), which is through the center and cuts in half the (straight-line), which is not through the center, also cuts () at right-angles. ἈλλὰδὴἡΔτὴνπρὸςὀρθὰςτεμνέτω λέγω,ὅτι nd so let cut at right-angles. I say that it καὶδίχααὐτὴντέμνει,τουτέστιν,ὅτιἴσηἐστὶνἡτῇ. also cuts () in half. That is to say, that is equal to Τῶν γὰρ αὐτῶν κατασκευασθέντων, ἐπεὶ ἴση ἐστὶν ἡ. τῇ,ἴσηἐστὶκαὶγωνίαἡὑπὸτῇὑπὸ. or, with the same construction, since is equal ἐστὶδὲκαὶὀρθὴἡὑπὸὀρθῇτῇὑπὸἴση δύο to, angle is also equal to [Prop. 1.5]. ἄρατρίγωνάἐστι,τὰςδύογωνίαςδυσὶγωνίαις nd the right-angle is also equal to the right-angle ἴσαςἔχοντακαὶμίανπλευρὰνμιᾷπλευρᾷἴσηνκοινὴναὐτῶν. Thus, and are two triangles having τὴνὑποτείνουσανὑπὸμίαντῶνἴσωνγωνιῶν καὶτὰς two angles equal to two angles, and one side equal to λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξει ἴση ἄρα one side (namely), their common (side), subtend- ἡτῇ. ing one of the equal angles. Thus, they will also have the ὰν ἄρα ἐν κύκλῳ εὐθεῖά τις διὰ τοῦ κέντρου εὐθεῖάν remaining sides equal to the (corresponding) remaining τινα μὴ διὰ τοῦ κέντρου δίχα τέμνῃ, καὶ πρὸς ὀρθὰς αὐτὴν sides [Prop. 1.26]. Thus, (is) equal to. τέμνει καὶ ἐὰν πρὸς ὀρθὰς αὐτὴν τέμνῃ, καὶ δίχα αὐτὴν Thus, in a circle, if any straight-line through the cenτέμνει ὅπερ ἔδει δεῖξαι. ter cuts in half any straight-line not through the center then it also cuts it at right-angles. nd (conversely) if it cuts it at right-angles then it also cuts it in half. (Which is) the very thing it was required to show.. Proposition 4 ὰν ἐν κύκλῳ δύο εὐθεῖαι τέμνωσιν ἀλλήλας μὴ δὶα τοῦ In a circle, if two straight-lines, which are not through κέντρου οὖσαι, οὐ τέμνουσιν ἀλλήλας δίχα. the center, cut one another then they do not cut one an- στω κύκλος ὁ Δ, καὶ ἐν αὐτῷ δύο εὐθεῖαι αἱ, other in half. Δ τεμνέτωσαν ἀλλήλας κατὰ τὸ μὴ διὰ τοῦ κέντρου Let be a circle, and within it, let two straightοὖσαι λέγω,ὅτιοὐτέμνουσινἀλλήλαςδίχα. lines, and, which are not through the center, cut ἰγὰρδυνατόν,τεμνέτωσανἀλλήλαςδίχαὥστεἴσην one another at (point). I say that they do not cut one εἶναιτὴνμὲντῇ,τὴνδὲτῇδ καὶεἰλήφθωτὸ another in half. κέντρον τοῦ Δ κύκλου, καὶ ἔστω τὸ, καὶ ἐπεζεύχθω or, if possible, let them cut one another in half, such ἡ. that is equal to, and to. nd let the πεὶοὖνεὐθεῖάτιςδιὰτοῦκέντρουἡεὐθεῖάντινα center of the circle have been found [Prop. 3.1], μὴ διὰ τοῦ κέντρου τὴν δίχα τέμνει, καὶ πρὸς ὀρθὰς and let it be (at point), and let have been joined. αὐτὴν τέμνει ὀρθὴ ἄρα ἐστὶν ἡ ὑπὸ πάλιν, ἐπεὶ εὐθεῖά Therefore, since some straight-line through the center, τιςἡεὐθεῖάντινατὴνδδίχατέμνει,καὶπρὸςὀρθὰς, cuts in half some straight-line not through the cenαὐτὴντέμνει ὀρθὴἄραἡὑπὸ.ἐδείχθηδὲκαὶἡὑπὸ ter,, it also cuts it at right-angles [Prop. 3.3]. Thus, ὀρθή ἴσηἄραἡὑπὸτῇὑπὸἡἐλάττωντῇ is a right-angle. gain, since some straight-line 73
μείζονι ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα αἱ, Δ τέμνουσιν cuts in half some straight-line, it also cuts it at right- ἀλλήλας δίχα. angles [Prop. 3.3]. Thus, (is) a right-angle. ut was also shown (to be) a right-angle. Thus, (is) equal to, the lesser to the greater. The very thing is impossible. Thus, and do not cut one another in half. ὰνἄραἐνκύκλῳδύοεὐθεῖαιτέμνωσινἀλλήλαςμὴδὶα Thus, in a circle, if two straight-lines, which are not τοῦ κέντρου οὖσαι, οὐ τέμνουσιν ἀλλήλας δίχα ὅπερ ἔδει through the center, cut one another then they do not cut δεῖξαι. one another in half. (Which is) the very thing it was required to show.. Proposition 5 ὰν δύο κύκλοι τέμνωσιν ἀλλήλους, οὐκ ἔσται αὐτῶν If two circles cut one another then they will not have τὸ αὐτὸ κέντρον. the same center. Δύογὰρκύκλοιοἱ,Δτεμνέτωσανἀλλήλους or let the two circles and cut one another κατὰ τὰ, σημεῖα. λέγω, ὅτι οὐκ ἔσται αὐτῶν τὸ αὐτὸ at points and. I say that they will not have the same κέντρον. center. ἰγὰρδυνατόν,ἔστωτὸ,καὶἐπεζεύχθωἡ,καὶ or, if possible, let be (the common center), and διήχθωἡ,ὡςἔτυχεν.καὶἐπεὶτὸσημεῖονκέντρον let have been joined, and let have been drawn ἐστὶτοῦκύκλου,ἵσηἐστὶνἡτῇ.πάλιν,ἐπεὶτὸ through (the two circles), at random. nd since point σημεῖονκέντρονἐστὶτοῦδκύκλου,ἴσηἐστὶνἡ is the center of the circle, is equal to. τῇ ἐδείχθη δὲ ἡ καὶ τῇ ἴση καὶ ἡ ἄρα τῇ gain, since point is the center of the circle, ἐστινἴσηἡἐλάσσωντῇμείζονι ὅπερἐστὶνἀδύνατον.οὐκ is equal to. ut was also shown (to be) equal ἄρατὸσημεῖονκέντρονἐστὶτῶν,δκύκλων. to. Thus, is also equal to, the lesser to the ὰν ἄρα δύο κύκλοι τέμνωσιν ἀλλήλους, οὐκ ἔστιν greater. The very thing is impossible. Thus, point is not 74
αὐτῶντὸαὐτὸκέντρον ὅπερἔδειδεῖξαι. the (common) center of the circles and. Thus, if two circles cut one another then they will not have the same center. (Which is) the very thing it was required to show.. Proposition 6 ὰν δύο κύκλοι ἐφάπτωνται ἀλλήλων, οὐκ ἔσται αὐτῶν If two circles touch one another then they will not τὸαὐτὸκέντρον. have the same center. Δύογὰρκύκλοιοἱ,Δἐφαπτέσθωσανἀλλήλων or let the two circles and touch one anκατὰτὸσημεῖον λέγω, ὅτιοὐκἔσταιαὐτῶντὸαὐτὸ other at point. I say that they will not have the same κέντρον. center. ἰγὰρδυνατόν,ἔστωτὸ,καὶἐπεζεύχθωἡ,καὶ or, if possible, let be (the common center), and διήχθω, ὡς ἔτυχεν, ἡ. let have been joined, and let have been drawn πεὶ οὖν τὸ σημεῖον κέντρον ἐστὶ τοῦ κύκλου, through (the two circles), at random. ἴσηἐστὶνἡτῇ.πάλιν,ἐπεὶτὸσημεῖονκέντρον Therefore, since point is the center of the circle ἐστὶτοῦδκύκλου,ἴσηἐστὶνἡτῇ.ἐδείχθηδὲἡ, is equal to. gain, since point is the τῇἴση καὶἡἄρατῇἐστινἴση,ἡἐλάττων center of the circle, is equal to. ut τῇμείζονι ὅπερἐστὶνἀδύνατον. οὐκἄρατὸσημεῖον was shown (to be) equal to. Thus, is also equal κέντρον ἐστὶ τῶν, Δ κύκλων. to, the lesser to the greater. The very thing is impos- ὰν ἄρα δύο κύκλοι ἐφάπτωνται ἀλλήλων, οὐκ ἔσται sible. Thus, point is not the (common) center of the αὐτῶν τὸ αὐτὸ κέντρον ὅπερ ἔδει δεῖξαι. circles and. Thus, if two circles touch one another then they will not have the same center. (Which is) the very thing it was required to show. Þ. Proposition 7 ὰνκύκλουἐπὶτῆςδιαμέτρουληφθῇτισημεῖον,ὃμή If some point, which is not the center of the circle, ἐστικέντροντοῦκύκλου,ἀπὸδὲτοῦσημείουπρὸςτὸν is taken on the diameter of a circle, and some straightκύκλον προσπίπτωσιν εὐθεῖαί τινες, μεγίστη μὲν ἔσται, ἐφ lines radiate from the point towards the (circumference ἧςτὸκέντρον,ἐλαχίστηδὲἡλοιπή,τῶνδὲἄλλωνἀεὶἡ of the) circle, then the greatest (straight-line) will be that ἔγγιον τῆς δὶα τοῦ κέντρου τῆς ἀπώτερον μείζων ἐστίν, on which the center (lies), and the least the remainder δύο δὲ μόνον ἴσαι ἀπὸ τοῦ σημείου προσπεσοῦνται πρὸς (of the same diameter). nd for the others, a (straightτὸνκύκλονἐφ ἑκάτερατῆςἐλαχίστης. line) nearer to the (straight-line) through the center is always greater than a (straight-line) further away. nd only two equal (straight-lines) will radiate from the point towards the (circumference of the) circle, (one) on each 75
(side) of the least (straight-line). Κ Θ στωκύκλοςὁδ,διάμετροςδὲαὐτοῦἔστωἡδ, Let be a circle, and let be its diameter, and καὶἐπὶτῆςδεἰλήφθωτισημεῖοντὸ,ὃμήἐστικέντρον let some point, which is not the center of the circle, τοῦκύκλου,κέντρονδὲτοῦκύκλουἔστωτὸ,καὶἀπὸτοῦ have been taken on. Let be the center of the circle. πρὸςτὸνδκύκλονπροσπιπτέτωσανεὐθεῖαίτινεςαἱ nd let some straight-lines,,, and, radiate,, λέγω, ὅτι μεγίστη μέν ἐστιν ἡ, ἐλαχίστη from towards (the circumference of) circle. I δὲἡδ,τῶνδὲἄλλωνἡμὲντῆςμείζων,ἡδὲ say that is the greatest (straight-line), the least, τῆς. and of the others, (is) greater than, and than πεζεύχθωσανγὰραἱ,,.καὶἐπεὶπαντὸς. τριγώνου αἱ δύο πλευραὶ τῆς λοιπῆς μείζονές εἰσιν, αἱ ἄρα or let,, and have been joined. nd since,τῆςμείζονέςεἰσιν.ἴσηδὲἡτῇ[αἱἄρα for every triangle (any) two sides are greater than the,ἴσαιεἰσὶτῇ] μείζωνἄραἡτῆς.πάλιν, remaining (side) [Prop. 1.20], and is thus greater ἐπεὶἴσηἐστὶνἡτῇ,κοινὴδὲἡ,δύοδὴαἱ, than. nd (is) equal to [thus, and δυσὶταῖς,ἴσαιεἰσίν.ἀλλὰκαὶγωνίαἡὑπὸ is equal to ]. Thus, (is) greater than. gain, γωνίαςτῆςὑπὸμείζων βάσιςἄραἡβάσεωςτῆς since is equal to, and (is) common, the two μείζωνἐστίν. διὰτὰαὐτὰδὴκαὶἡτῆςμείζων (straight-lines), are equal to the two (straight- ἐστίν. lines), (respectively). ut, angle (is) also Πάλιν,ἐπεὶαἱ,τῆςμείζονέςεἰσιν,ἴσηδὲἡ greater than angle. Thus, the base is greater τῇ Δ, αἱ ἄρα, τῆς Δ μείζονές εἰσιν. κοινὴ than the base. Thus, the base is greater than the ἀφῃρήσθωἡ λοιπὴἄραἡλοιπῆςτῆςδμείζων base [Prop. 1.24]. So, for the same (reasons), is ἐστίν.μεγίστημὲνἄραἡ,ἐλαχίστηδὲἡδ,μείζωνδὲ also greater than. ἡ μὲν τῆς, ἡ δὲ τῆς. gain, since and are greater than Λέγω,ὅτικαὶἀπὸτοῦσημείουδύομόνονἴσαιπρο- [Prop. 1.20], and (is) equal to, and σπεσοῦνται πρὸς τὸν Δ κύκλον ἐφ ἑκάτερα τῆς Δ are thus greater than. Let have been taken from ἐλαχίστης. συνεστάτω γὰρ πρὸς τῇ εὐθείᾳ καὶ τῷ πρὸς both. Thus, the remainder is greater than the reαὐτῇσημείῳτῷτῇὑπὸγωνίᾳἴσηἡὑπὸθ,καὶ mainder. Thus, (is) the greatest (straight-line), ἐπεζεύχθωἡθ.ἐπεὶοὖνἴσηἐστὶνἡτῇθ,κοινὴ the least, and (is) greater than, and than δὲἡ,δύοδὴαἱ,δυσὶταῖςθ,ἴσαιεἰσίν. καὶγωνίαἡὑπὸγωνίᾳτῇὑπὸθἴση βάσιςἄρα I also say that from point only two equal (straight- ἡβάσειτῇθἴσηἐστίν. λέγωδή,ὅτιτῇἄλλη lines) will radiate towards (the circumference of) circle ἴση οὐ προσπεσεῖται πρὸς τὸν κύκλον ἀπὸ τοῦ σημείου., (one) on each (side) of the least (straight-line) εἰγὰρδυνατόν,προσπιπτέτωἡκ.καὶἐπεὶἡκτῇ. or let the (angle) H, equal to angle, have ἴσηἐστίν,ἀλλὰἡθτῇ[ἴσηἐστίν],καὶἡκἄρατῇ been constructed on the straight-line, at the point Θἐστινἴση,ἡἔγγιοντῆςδιὰτοῦκέντρουτῇἀπώτερον on it [Prop. 1.23], and let H have been joined. There- ἴση ὅπερἀδύνατον. οὐκἄραἀπὸτοῦσημείουἑτέρατις fore, since is equal to H, and (is) common, K H 76
προσπεσεῖταιπρὸςτὸνκύκλονἴσητῇ μίαἄραμόνη. the two (straight-lines), are equal to the two ὰν ἄρα κύκλου ἐπὶ τῆς διαμέτρου ληφθῇ τι σημεῖον, (straight-lines) H, (respectively). nd angle ὃμήἐστικέντροντοῦκύκλου,ἀπὸδὲτοῦσημείουπρὸς (is) equal to angle H. Thus, the base is equal to τὸν κύκλον προσπίπτωσιν εὐθεῖαί τινες, μεγίστη μὲν ἔσται, the base H [Prop. 1.4]. So I say that another (straight- ἐφ ἧςτὸκέντρον,ἐλαχίστηδὲἡλοιπή,τῶνδὲἄλλωνἀεὶἡ line) equal to will not radiate towards (the circumfer- ἔγγιοντῆςδὶατοῦκέντρουτῆςἀπώτερονμείζωνἐστίν,δύο ence of) the circle from point. or, if possible, let K δὲμόνονἴσαιἀπὸτοῦαὐτοῦσημείουπροσπεσοῦνταιπρὸς (so) radiate. nd since K is equal to, but H [is τὸνκύκλονἐφ ἑκάτερατῆςἐλαχίστης ὅπερἔδειδεῖξαι. equal] to, K is thus also equal to H, the nearer to the (straight-line) through the center equal to the further away. The very thing (is) impossible. Thus, another (straight-line) equal to will not radiate from the point towards (the circumference of) the circle. Thus, (there is) only one (such straight-line). Thus, if some point, which is not the center of the circle, is taken on the diameter of a circle, and some straight-lines radiate from the point towards the (circumference of the) circle, then the greatest (straight-line) will be that on which the center (lies), and the least the remainder (of the same diameter). nd for the others, a (straight-line) nearer to the (straight-line) through the center is always greater than a (straight-line) further away. nd only two equal (straight-lines) will radiate from the same point towards the (circumference of the) circle, (one) on each (side) of the least (straight-line). (Which is) the very thing it was required to show. Presumably, in an angular sense. This is not proved, except by reference to the figure.. Proposition 8 ὰν κύκλου ληφθῇ τι σημεῖον ἐκτός, ἀπὸ δὲ τοῦ If some point is taken outside a circle, and some σημείου πρὸς τὸν κύκλον διαχθῶσιν εὐθεῖαί τινες, ὧν μία straight-lines are drawn from the point to the (circumμὲν διὰ τοῦ κέντρου, αἱ δὲ λοιπαί, ὡς ἔτυχεν, τῶν μὲν πρὸς ference of the) circle, one of which (passes) through τὴν κοίλην περιφέρειαν προσπιπτουσῶν εὐθειῶν μεγίστη the center, the remainder (being) random, then for the μέν ἐστιν ἡ διὰ τοῦ κέντρου, τῶν δὲ ἄλλων ἀεὶ ἡ ἔγγιον τῆς straight-lines radiating towards the concave (part of the) διὰ τοῦ κέντρου τῆς ἀπώτερον μείζων ἐστίν, τῶν δὲ πρὸς circumference, the greatest is that (passing) through the τὴνκυρτὴνπεριφέρειανπροσπιπτουσῶνεὐθειῶνἐλαχίστη center. or the others, a (straight-line) nearer to the μέν ἐστιν ἡ μεταξὺ τοῦ τε σημείου καὶ τῆς διαμέτρου, τῶν (straight-line) through the center is always greater than δὲ ἄλλων ἀεὶ ἡ ἔγγιον τῆς ἐλαχίστης τῆς ἀπώτερόν ἐστιν one further away. or the straight-lines radiating towards ἐλάττων, δύο δὲ μόνον ἴσαι ἀπὸ τοῦ σημείου προσπεσοῦνται the convex (part of the) circumference, the least is that πρὸς τὸν κύκλον ἐφ ἑκάτερα τῆς ἐλαχίστης. between the point and the diameter. or the others, a στω κύκλος ὁ, καὶ τοῦ εἰλήφθω τι σημεῖον (straight-line) nearer to the least (straight-line) is always ἐκτὸς τὸ Δ, καὶ ἀπ αὐτοῦ διήχθωσαν εὐθεῖαί τινες αἱ Δ, less than one further away. nd only two equal (straight- Δ, Δ, Δ, ἔστω δὲ ἡ Δ διὰ τοῦ κέντρου. λέγω, lines) will radiate from the point towards the (circum- ὅτι τῶν μὲν πρὸς τὴν κοίλην περιφέρειαν προσπι- ference of the) circle, (one) on each (side) of the least πτουσῶν εὐθειῶν μεγίστη μέν ἐστιν ἡ διὰ τοῦ κέντρου ἡ (straight-line). Δ,μείζωνδὲἡμὲνΔτῆςΔἡδὲΔτῆςΔ,τῶν Let be a circle, and let some point have been δὲ πρὸς τὴν ΘΛΚ κυρτὴν περιφέρειαν προσπιπτουσῶν taken outside, and from it let some straight-lines, εὐθειῶνἐλαχίστημένἐστινἡδἡμεταξὺτοῦσημείουκαὶ,,, and, have been drawn through (the τῆςδιαμέτρουτῆς,ἀεὶδὲἡἔγγιοντῆςδἐλαχίστης circle), and let be through the center. I say that for ἐλάττωνἐστὶτῆςἀπώτερον,ἡμὲνδκτῆςδλ,ἡδὲδλ the straight-lines radiating towards the concave (part of 77
τῆς ΔΘ. the) circumference,, the greatest is the one (passing) through the center, (namely), and (that) (is) greater than, and than. or the straight-lines radiating towards the convex (part of the) circumference, HLK, the least is the one between the point and the diameter, (namely), and a (straight-line) nearer to the least (straight-line) is always less than one farther away, (so that) K (is less) than L, and L than than H. Θ Λ Κ H L K N Μ Ν M ἰλήφθω γὰρ τὸ κέντρον τοῦ κύκλου καὶ ἔστω τὸ or let the center of the circle have been found Μ καὶἐπεζεύχθωσαναἱμ,μ,μ,μκ,μλ,μθ. [Prop. 3.1], and let it be (at point) M [Prop. 3.1]. nd let ΚαὶἐπεὶἴσηἐστὶνἡΜτῇΜ,κοινὴπροσκείσθωἡ M, M, M, MK, ML, and MH have been joined. ΜΔ ἡἄραδἴσηἐστὶταῖςμ,μδ.ἀλλ αἱμ,μδ nd since M is equal to M, let M have been τῆςδμείζονέςεἰσιν καὶἡδἄρατῆςδμείζωνἐστίν. added to both. Thus, is equal to M and M. ut, πάλιν,ἐπεὶἴσηἐστὶνἡμτῇμ,κοινὴδὲἡμδ,αἱμ, M and M is greater than [Prop. 1.20]. Thus, ΜΔἄραταῖςΜ,ΜΔἴσαιεἰσίν καὶγωνίαἡὑπὸμδ is also greater than. gain, since M is equal γωνίαςτῆςὑπὸμδμείζωνἐστίν.βάσιςἄραἡδβάσεως to M, and M (is) common, the (straight-lines) M, τῆςδμείζωνἐστίν ὁμοίωςδὴδείξομεν,ὅτικαὶἡδτῆς M are thus equal to M, M. nd angle M is Δμείζωνἐστίν μεγίστημὲνἄραἡδ,μείζωνδὲἡμὲν greater than angle M. Thus, the base is greater ΔτῆςΔ,ἡδὲΔτῆςΔ. than the base [Prop. 1.24]. So, similarly, we can ΚαὶἐπεὶαἱΜΚ,ΚΔτῆςΜΔμείζονέςεἰσιν,ἴσηδὲἡ show that is also greater than. Thus, (is) the ΜτῇΜΚ,λοιπὴἄραἡΚΔλοιπῆςτῆςΔμείζωνἐστίν greatest (straight-line), and (is) greater than, ὥστεἡδτῆςκδἐλάττωνἐστίν καὶἐπεὶτριγώνουτοῦ and than. ΜΛΔἐπὶμιᾶςτῶνπλευρῶντῆςΜΔδύοεὐθεῖαιἐντὸς nd since MK and K is greater than M [Prop. συνεστάθησαναἱμκ,κδ,αἱἄραμκ,κδτῶνμλ,λδ 1.20], and M (is) equal to MK, the remainder K ἐλάττονέςεἰσιν ἴσηδὲἡμκτῇμλ λοιπὴἄραἡδκ is thus greater than the remainder. So is less λοιπῆςτῆςδλἐλάττωνἐστίν. ὁμοίωςδὴδείξομεν, ὅτι than K. nd since in triangle ML, the two interκαὶἡδλτῆςδθἐλάττωνἐστίν ἐλαχίστημὲνἄραἡδ, nal straight-lines MK and K were constructed on one ἐλάττωνδὲἡμὲνδκτῆςδλἡδὲδλτῆςδθ. of the sides, M, then MK and K are thus less than Λέγω, ὅτι καὶ δύο μόνον ἴσαι ἀπὸ τοῦ Δ σημείου ML and L [Prop. 1.21]. nd MK (is) equal to ML. προσπεσοῦνται πρὸς τὸν κύκλον ἐφ ἑκάτερα τῆς Δ Thus, the remainder K is less than the remainder L. ἐλαχίστης συνεστάτω πρὸς τῇ ΜΔ εὐθείᾳ καὶ τῷ πρὸς So, similarly, we can show that L is also less than H. αὐτῇσημείῳτῷμτῇὑπὸκμδγωνίᾳἴσηγωνίαἡὑπὸ Thus, (is) the least (straight-line), and K (is) less ΔΜ,καὶἐπεζεύχθωἡΔ.καὶἐπεὶἴσηἐστὶνἡΜΚτῇ than L, and L than H. Μ,κοινὴδὲἡΜΔ,δύοδὴαἱΚΜ,ΜΔδύοταῖςΜ,ΜΔ I also say that only two equal (straight-lines) will radi- 78
ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ καὶ γωνία ἡ ὑπὸ ΚΜΔ γωνίᾳ ate from point towards (the circumference of) the cirτῇ ὑπὸ ΜΔ ἴση βάσις ἄρα ἡ ΔΚ βάσει τῇ Δ ἴση ἐστίν. cle, (one) on each (side) on the least (straight-line),. λέγω [δή], ὅτι τῇ ΔΚ εὐθείᾳἄλληἴση οὐπροσπεσεῖται Let the angle M, equal to angle KM, have been πρὸςτὸνκύκλονἀπὸτοῦδσημείου.εἰγὰρδυνατόν,προ- constructed on the straight-line M, at the point M on σπιπτέτωκαὶἔστωἡδν.ἐπεὶοὖνἡδκτῇδνἐστινἴση, it [Prop. 1.23], and let have been joined. nd since ἀλλ ἡδκτῇδἐστινἴση,καὶἡδἄρατῇδνἐστιν MK is equal to M, and M (is) common, the two ἴση, ἡ ἔγγιον τῆς Δ ἐλαχίστης τῇ ἀπώτερον[ἐστιν] ἴση (straight-lines) KM, M are equal to the two (straight- ὅπερ ἀδύνατον ἐδείχθη. οὐκ ἄρα πλείους ἢ δύο ἴσαι πρὸς lines) M, M, respectively. nd angle KM (is) τὸνκύκλονἀπὸτοῦδσημείουἐφ ἑκάτερατῆςδ equal to angle M. Thus, the base K is equal to the ἐλαχίστης προσπεσοῦνται. base [Prop. 1.4]. [So] I say that another (straight- ὰνἄρακύκλουληφθῇτισημεῖονἐκτός,ἀπὸδὲτοῦ line) equal to K will not radiate towards the (circumσημείου πρὸς τὸν κύκλον διαχθῶσιν εὐθεῖαί τινες, ὧν μία ference of the) circle from point. or, if possible, let μὲνδιὰτοῦκέντρουαἱδὲλοιπαί,ὡςἔτυχεν,τῶνμὲνπρὸς (such a straight-line) radiate, and let it be N. Thereτὴν κοίλην περιφέρειαν προσπιπτουσῶν εὐθειῶν μεγίστη fore, since K is equal to N, but K is equal to, μένἐστινἡδιὰτοῦκέντου,τῶνδὲἄλλωνἀεὶἡἔγγιοντῆς then is thus also equal to N, (so that) a (straightδιὰ τοῦ κέντρου τῆς ἀπώτερον μείζων ἐστίν, τῶν δὲ πρὸς line) nearer to the least (straight-line) [is] equal to τὴν κυρτὴν περιφέρειαν προσπιπτουσῶν εὐθειῶν ἐλαχίστη one further away. The very thing was shown (to be) imμέν ἐστιν ἡ μεταξὺ τοῦ τε σημείου καὶ τῆς διαμέτρου, τῶν possible. Thus, not more than two equal (straight-lines) δὲ ἄλλων ἀεὶ ἡ ἔγγιον τῆς ἐλαχίστης τῆς ἀπώτερόν ἐστιν will radiate towards (the circumference of) circle ἐλάττων, δύο δὲ μόνον ἴσαι ἀπὸ τοῦ σημείου προσπεσοῦνται from point, (one) on each side of the least (straightπρὸς τὸν κύκλον ἐφ ἑκάτερα τῆς ἐλαχίστης ὅπερ ἔδει line). δεῖξαι. Thus, if some point is taken outside a circle, and some straight-lines are drawn from the point to the (circumference of the) circle, one of which (passes) through the center, the remainder (being) random, then for the straightlines radiating towards the concave (part of the) circumference, the greatest is that (passing) through the center. or the others, a (straight-line) nearer to the (straightline) through the center is always greater than one further away. or the straight-lines radiating towards the convex (part of the) circumference, the least is that between the point and the diameter. or the others, a (straight-line) nearer to the least (straight-line) is always less than one further away. nd only two equal (straightlines) will radiate from the point towards the (circumference of the) circle, (one) on each (side) of the least (straight-line). (Which is) the very thing it was required to show. Presumably, in an angular sense. This is not proved, except by reference to the figure.. Proposition 9 ὰν κύκλου ληφθῇ τι σημεῖον ἐντός, ἀπο δὲ τοῦ If some point is taken inside a circle, and more than σημείου πρὸς τὸν κύκλον προσπίπτωσι πλείους ἢ δύο ἴσαι two equal straight-lines radiate from the point towards εὐθεῖαι, τὸ ληφθὲν σημεῖον κέντρον ἐστὶ τοῦ κύκλου. the (circumference of the) circle, then the point taken is στω κύκλος ὁ, ἐντὸς δὲ αὐτοῦ σημεῖον τὸ Δ, καὶ the center of the circle. ἀπὸτοῦδπρὸςτὸνκύκλονπροσπιπτέτωσανπλείους Let be a circle, and a point inside it, and let ἢδύοἴσαιεὐθεῖαιαἱδ,δ,δ λέγω,ὅτιτὸδσημεῖον more than two equal straight-lines,,, and, raκέντρον ἐστὶ τοῦ κύκλου. diate from towards (the circumference of) circle. 79
Λ I say that point is the center of circle. L Κ K Θ H πεζεύχθωσαν γὰρ αἱ, καὶ τετμήσθωσαν or let and have been joined, and (then) δίχακατὰτὰ,σημεῖα,καὶἐπιζευχθεῖσαιαἱδ,δ have been cut in half at points and (respectively) διήχθωσανἐπὶτὰ,κ,θ,λσημεῖα. [Prop. 1.10]. nd and being joined, let them πεὶ οὖν ἴση ἐστὶν ἡ τῇ, κοινὴ δὲ ἡ Δ, δύο have been drawn through to points, K, H, and L. δὴαἱ,δδύοταῖς,δἴσαιεἰσίν καὶβάσιςἡδ Therefore, since is equal to, and (is) comβάσειτῇδἴση γωνίαἄραἡὑπὸδγωνίᾳτῇὑπὸδ mon, the two (straight-lines), are equal to the ἴση ἐστίν ὀρθὴ ἄρα ἑκατέρα τῶν ὑπὸ Δ, Δ γωνιῶν ἡ two (straight-lines), (respectively). nd the base Κἄρατὴντέμνειδίχακαὶπρὸςὀρθάς.καὶἐπεί,ἐὰν (is) equal to the base. Thus, angle is equal ἐν κύκλῳ εὐθεῖά τις εὐθεῖάν τινα δίχα τε καὶ πρὸς ὀρθὰς to angle [Prop. 1.8]. Thus, angles and τέμνῃ, ἐπὶ τῆς τεμνούσης ἐστὶ τὸ κέντρον τοῦ κύκλου, ἐπὶ (are) each right-angles [ef. 1.10]. Thus, K cuts in τῆςκἄραἐστὶτὸκέντροντοῦκύκλου.διὰτὰαὐτὰδὴκαὶ half, and at right-angles. nd since, if some straight-line ἐπὶτῆςθλἐστιτὸκέντροντοῦκύκλου. καὶοὐδὲν in a circle cuts some (other) straight-line in half, and at ἕτερονκοινὸνἔχουσιναἱκ,θλεὐθεῖαιἢτὸδσημεῖον right-angles, then the center of the circle is on the former τὸ Δ ἄρα σημεῖον κέντρον ἐστὶ τοῦ κύκλου. (straight-line) [Prop. 3.1 corr.], the center of the circle is ὰνἄρακύκλουληφθῇτισημεῖονἐντός,ἀπὸδὲτοῦ thus on K. So, for the same (reasons), the center of σημείου πρὸς τὸν κύκλον προσπίπτωσι πλείους ἢ δύο ἴσαι circle is also on HL. nd the straight-lines K and εὐθεῖαι, τὸ ληφθὲν σημεῖον κέντρον ἐστὶ τοῦ κύκλου ὅπερ HL have no common (point) other than point. Thus, ἔδειδεῖξαι. point is the center of circle. Thus, if some point is taken inside a circle, and more than two equal straight-lines radiate from the point towards the (circumference of the) circle, then the point taken is the center of the circle. (Which is) the very thing it was required to show.. Proposition 10 Κύκλοςκύκλονοὐτέμνεικατὰπλείονασημεῖαἢδύο. circle does not cut a(nother) circle at more than two ἰ γὰρ δυνατόν, κύκλος ὁ κύκλον τὸν Δ points. τεμνέτω κατὰ πλείονα σημεῖα ἢ δύο τὰ,,, Θ, καὶ or, if possible, let the circle cut the circle ἐπιζευχθεῖσαιαἱθ,δίχατεμνέσθωσανκατὰτὰκ,λ at more than two points,,,, and H. nd H and σημεῖα καὶἀπὸτῶνκ,λταῖςθ,πρὸςὀρθὰςἀχθεῖσαι being joined, let them (then) have been cut in half αἱκ,λμδιήχθωσανἐπὶτὰ,σημεῖα. at points K and L (respectively). nd K and LM being drawn at right-angles to H and from K and L (respectively) [Prop. 1.11], let them (then) have been drawn through to points and (respectively). 80
Μ Θ Ν Κ Ο Λ Ξ πεὶ οὖν ἐν κύκλῳ τῷ εὐθεῖά τις ἡ εὐθεῖάν Therefore, since in circle some straight-line τινατὴνθδίχακαὶπρὸςὀρθὰςτέμνει,ἐπὶτῆςἄρα cuts some (other) straight-line H in half, and at ἐστὶτὸκέντροντοῦκύκλου.πάλιν,ἐπεὶἐνκύκλῳτῷ right-angles, the center of circle is thus on αὐτῷτῷεὐθεῖάτιςἡνξεὐθεῖάντινατὴνδίχα [Prop. 3.1 corr.]. gain, since in the same circle καὶ πρὸς ὀρθὰς τέμνει, ἐπὶ τῆς ΝΞ ἄρα ἐστὶ τὸ κέντρον some straight-line N O cuts some (other straight-line) τοῦκύκλου. ἐδείχθηδὲκαὶἐπὶτῆς,καὶκατ in half, and at right-angles, the center of circle is οὐδὲνσυμβάλλουσιναἱ,νξεὐθεῖαιἢκατὰτὸο τὸ thus on NO [Prop. 3.1 corr.]. nd it was also shown (to Οἄρασημεῖονκέντρονἐστὶτοῦκύκλου. ὁμοίωςδὴ be) on. nd the straight-lines and NO meet at δείξομεν,ὅτικαὶτοῦδκύκλουκέντρονἐστὶτὸο δύο no other (point) than P. Thus, point P is the center of ἄρα κύκλων τεμνόντων ἀλλήλους τῶν, Δ τὸ αὐτό circle. So, similarly, we can show that P is also the ἐστι κέντρον τὸ Ο ὅπερ ἐστὶν ἀδύνατον. center of circle. Thus, two circles cutting one an- Οὐκἄρακύκλοςκύκλοντέμνεικατὰπλείονασημεῖαἢ other, and, have the same center P. The very δύο ὅπερἔδειδεῖξαι. thing is impossible [Prop. 3.5]. Thus, a circle does not cut a(nother) circle at more than two points. (Which is) the very thing it was required to show.. Proposition 11 ὰν δύο κύκλοι ἐφάπτωνται ἀλλήλων ἐντός, καὶ ληφθῇ If two circles touch one another internally, and their αὐτῶν τὰ κέντρα, ἡ ἐπὶ τὰ κέντρα αὐτῶν ἐπιζευγνυμένη centers are found, then the straight-line joining their cenεὐθεῖα καὶ ἐκβαλλομένη ἐπὶ τὴν συναφὴν πεσεῖται τῶν ters, being produced, will fall upon the point of union of κύκλων. the circles. Δύο γὰρ κύκλοι οἱ, Δ ἐφαπτέσθωσαν ἀλλήλων or let two circles, and, touch one another ἐντὸςκατὰτὸσημεῖον,καὶεἰλήφθωτοῦμὲνκύκλου internally at point, and let the center of circle κέντροντὸ,τοῦδὲδτὸ λέγω,ὅτιἡἀπὸτοῦἐπὶ have been found [Prop. 3.1], and (the center) of (cirτὸ ἐπιζευγνυμένη εὐθεῖα ἐκβαλλομένη ἐπὶ τὸ πεσεῖται. cle) [Prop. 3.1]. I say that the straight-line joining Μὴ γάρ, ἀλλ εἰ δυνατόν, πιπτέτω ὡς ἡ Θ, καὶ to, being produced, will fall on. ἐπεζεύχθωσαναἱ,. or (if) not then, if possible, let it fall like H (in πεὶ οὖν αἱ, τῆς, τουτέστι τῆς Θ, μείζονές the figure), and let and have been joined. εἰσιν, κοινὴ ἀφῃρήσθω ἡ λοιπὴ ἄρα ἡ λοιπῆς τῆς Therefore, since and is greater than, that Θμείζωνἐστίν. ἴσηδὲἡτῇδ καὶἡδἄρα is to say H [Prop. 1.20], let have been taken from τῆς Θ μείζων ἐστὶν ἡ ἐλάττων τῆς μείζονος ὅπερ ἐστὶν both. Thus, the remainder is greater than the re- ἀδύνατον οὐκἄραἡἀπὸτοῦἐπὶτὸἐπιζευγνυμένη mainder H. nd (is) equal to. Thus, is εὐθεὶα ἐκτὸς πεσεῖται κατὰ τὸ ἄρα ἐπὶ τῆς συναφῆς also greater than H, the lesser than the greater. The πεσεῖται. very thing is impossible. Thus, the straight-line joining to will not fall outside (one circle but inside the other). Thus, it will fall upon the point of union (of the circles) M H N K P L O 81
Θ at point. H ὰν ἄρα δύο κύκλοι ἐφάπτωνται ἀλλήλων ἐντός,[καὶ Thus, if two circles touch one another internally, [and ληφθῇ αὐτῶν τὰ κέντρα], ἡ ἐπὶ τὰ κέντρα αὐτῶν ἐπιζευ- their centers are found], then the straight-line joining γνυμένη εὐθεῖα[καὶ ἐκβαλλομένη] ἐπὶ τὴν συναφὴν πεσεῖται their centers, [being produced], will fall upon the point τῶνκύκλων ὅπερἔδειδεῖξαι. of union of the circles. (Which is) the very thing it was required to show.. Proposition 12 ὰν δύο κύκλοι ἐφάπτωνται ἀλλήλων ἐκτός, ἡ ἐπὶ τὰ If two circles touch one another externally then the κέντρα αὐτῶν ἐπιζευγνυμένη διὰ τῆς ἐπαφῆς ἐλεύσεται. (straight-line) joining their centers will go through the point of union. Δύογὰρκύκλοιοἱ,Δἐφαπτέσθωσανἀλλήλων or let two circles, and, touch one an- ἐκτὸς κατὰ τὸ σημεῖον, καὶ εἰλήφθω τοῦ μὲν other externally at point, and let the center of κέντροντὸ,τοῦδὲδτὸ λέγω, ὅτιἡἀπὸτοῦ have been found [Prop. 3.1], and (the center) of ἐπὶτὸἐπιζευγνυμένηεὐθεῖαδιὰτῆςκατὰτὸἐπαφῆς [Prop. 3.1]. I say that the straight-line joining to will ἐλεύσεται. go through the point of union at. Μὴγάρ, ἀλλ εἰδυνατόν, ἐρχέσθωὡςἡδ,καὶ or (if) not then, if possible, let it go like (in ἐπεζεύχθωσαναἱ,. the figure), and let and have been joined. πεὶοὖντὸσημεῖονκέντρονἐστὶτοῦκύκλου, Therefore, since point is the center of circle, ἴσηἐστὶνἡτῇ.πάλιν,ἐπεὶτὸσημεῖονκέντρον is equal to. gain, since point is the center of ἐστὶτοῦδκύκλου,ἴσηἐστὶνἡτῇδ.ἐδείχθη circle, is equal to. nd was also shown 82
δὲκαὶἡτῇἴση αἱἄρα,ταῖς,δἴσαι (to be) equal to. Thus, the (straight-lines) and εἰσίν ὥστεὅληἡτῶν,μείζωνἐστίν ἀλλὰκαὶ are equal to the (straight-lines) and. So the ἐλάττων ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἡ ἀπὸ τοῦ ἐπὶ whole of is greater than and. ut, (it is) also τὸ ἐπιζευγνυμένη εὐθεῖα διὰ τῆς κατὰ τὸ ἐπαφῆς οὐκ less [Prop. 1.20]. The very thing is impossible. Thus, the ἐλεύσεται δι αὐτῆς ἄρα. straight-line joining to cannot not go through the ὰν ἄρα δύο κύκλοι ἐφάπτωνται ἀλλήλων ἐκτός, ἡ ἐπὶ point of union at. Thus, (it will go) through it. τὰ κέντρα αὐτῶν ἐπιζευγνυμένη[εὐθεῖα] διὰ τῆς ἐπαφῆς Thus, if two circles touch one another externally then ἐλεύσεται ὅπερ ἔδει δεῖξαι. the [straight-line] joining their centers will go through the point of union. (Which is) the very thing it was required to show.. Proposition 13 Κύκλος κύκλου οὐκ ἐφάπτεται κατὰ πλείονα σημεῖα ἢ circle does not touch a(nother) circle at more than καθ ἕν, ἐάν τε ἐντὸς ἐάν τε ἐκτὸς ἐφάπτηται. one point, whether they touch internally or externally. Κ K Θ H ἰγὰρδυνατόν,κύκλοςὁδκύκλουτοῦδ or, if possible, let circle touch circle ἐφαπτέσθω πρότερον ἐντὸς κατὰ πλείονα σημεῖα ἢ ἓν τὰ Δ, first of all, internally at more than one point, and.. nd let the center of circle have been found Καὶ εἰλήφθω τοῦ μὲν Δ κύκλου κέντρον τὸ, τοῦ [Prop. 3.1], and (the center) H of [Prop. 3.1]. δὲδτὸθ. Thus, the (straight-line) joining and H will fall on ἄραἀπὸτοῦἐπὶτὸθἐπιζευγνυμένηἐπὶτὰ, and [Prop. 3.11]. Let it fall like H (in the Δπεσεῖται. πιπτέτωὡςἡθδ.καὶἐπεὶτὸσημεῖον figure). nd since point is the center of circle, κέντρονἐστὶτοῦδκύκλου,ἴσηἐστὶνἡτῇδ is equal to. Thus, (is) greater than H. μείζωνἄραἡτῆςθδ πολλῷἄραμείζωνἡθτῆςθδ. Thus, H (is) much greater than H. gain, since point πάλιν,ἐπεὶτὸθσημεῖονκέντρονἐστὶτοῦδκύκλου, H is the center of circle, H is equal to H. ἴσηἐστὶνἡθτῇθδ ἐδείχθηδὲαὐτῆςκαὶπολλῷμείζων ut it was also shown (to be) much greater than it. The ὅπερ ἀδύνατον οὐκ ἄρα κύκλος κύκλου ἐφάπτεται ἐντὸς very thing (is) impossible. Thus, a circle does not touch κατὰ πλείονα σημεῖα ἢ ἕν. a(nother) circle internally at more than one point. Λέγωδή,ὅτιοὐδὲἐκτός. So, I say that neither (does it touch) externally (at ἰ γὰρ δυνατόν, κύκλος ὁ Κ κύκλου τοῦ Δ more than one point). ἐφαπτέσθωἐκτὸςκατὰπλείονασημεῖαἢἓντὰ,,καὶ or, if possible, let circle K touch circle ἐπεζεύχθω ἡ. externally at more than one point, and. nd let πεὶ οὖν κύκλων τῶν Δ, Κ εἴληπται ἐπὶ τῆς have been joined. περιφερείαςἑκατέρουδύοτυχόντασημεῖατὰ,,ἡἐπὶ Therefore, since two points, and, have been taken τὰ σημεῖα ἐπιζευγνυμένη εὐθεῖα ἐντὸς ἑκατέρου πεσεῖται at random on the circumference of each of the circles ἀλλὰ τοῦ μὲν Δ ἐντὸς ἔπεσεν, τοῦ δὲ Κ ἐκτός and K, the straight-line joining the points will ὅπερ ἄτοπον οὐκ ἄρα κύκλος κύκλου ἐφάπτεται ἐκτὸς κατὰ fall inside each (circle) [Prop. 3.2]. ut, it fell inside πλείονα σημεῖα ἢ ἕν. ἐδείχθη δέ, ὅτι οὐδὲ ἐντός., and outside K [ef. 3.3]. The very thing 83
Κύκλος ἄρα κύκλου οὐκ ἐφάπτεται κατὰ πλείονα σημεῖα (is) absurd. Thus, a circle does not touch a(nother) circle ἢ[καθ ]ἕν,ἐάντεἐντὸςἐάντεἐκτὸςἐφάπτηται ὅπερἔδει externally at more than one point. nd it was shown that δεῖξαι. neither (does it) internally. Thus, a circle does not touch a(nother) circle at more than one point, whether they touch internally or externally. (Which is) the very thing it was required to show. The reek text has, which is obviously a mistake.. Proposition 14 ν κύκλῳ αἱ ἴσαι εὐθεῖαι ἴσον ἀπέχουσιν ἀπὸ τοῦ In a circle, equal straight-lines are equally far from the κέντρου, καὶ αἱ ἴσον ἀπέχουσαι ἀπὸ τοῦ κέντρου ἴσαι center, and (straight-lines) which are equally far from the ἀλλήλαιςεἰσίν. center are equal to one another. στωκύκλοςὁδ,καὶἐναὐτῷἴσαιεὐθεῖαιἔστω- Let be a circle, and let and be equal σαναἱ,δ λέγω,ὅτιαἱ,δἴσονἀπέχουσινἀπὸ straight-lines within it. I say that and are equally τοῦκέντρου. far from the center. ἰλήφθωγὰρτὸκέντοντοῦδκύκλουκαὶἔστωτὸ or let the center of circle have been found,καὶἀπὸτοῦἐπὶτὰς,δκάθετοιἤχθωσαναἱ, [Prop. 3.1], and let it be (at). nd let and, καὶ ἐπεζεύχθωσαν αἱ,. have been drawn from (point), perpendicular to πεὶ οὖν εὐθεῖά τις δὶα τοῦ κέντρου ἡ εὐθεῖάν τινα and (respectively) [Prop. 1.12]. nd let and μὴ διὰ τοῦ κέντρου τὴν πρὸς ὀρθὰς τέμνει, καὶ δίχα have been joined. αὐτὴντέμνει. ἴσηἄραἡτῇ διπλῆἄραἡτῆς Therefore, since some straight-line,, through the.διὰτὰαὐτὰδὴκαὶἡδτῆςἐστιδιπλῆ καίἐστιν center (of the circle), cuts some (other) straight-line,, ἴσηἡτῇδ ἴσηἄρακαὶἡτῇ.καὶἐπεὶἴσηἐστὶν not through the center, at right-angles, it also cuts it in ἡτῇ,ἴσονκαὶτὸἀπὸτῆςτῷἀπὸτῆς.ἀλλὰ half [Prop. 3.3]. Thus, (is) equal to. Thus, τῷμὲνἀπὸτῆςἴσατὰἀπὸτῶν, ὀρθὴγὰρἡ (is) double. So, for the same (reasons), is also πρὸςτῷγωνία τῷδὲἀπὸτῆςἴσατὰἀπὸτῶν, double. nd is equal to. Thus, (is) ὀρθὴγὰρἡπρὸςτῷγωνία τὰἄραἀπὸτῶν,ἴσα also equal to. nd since is equal to, the ἐστὶτοῖςἀπὸτῶν,,ὧντὸἀπὸτῆςἴσονἐστὶτῷ (square) on (is) also equal to the (square) on. ἀπὸτῆς ἴσηγάρἐστινἡτῇ λοιπὸνἄρατὸἀπὸ ut, the (sum of the squares) on and (is) equal τῆςτῷἀπὸτῆςἴσονἐστίν ἴσηἄραἡτῇ.ἐν to the (square) on. or the angle at (is) a rightδὲ κύκλῳ ἴσον ἀπέχειν ἀπὸ τοῦ κέντρου εὐθεῖαι λέγονται, angle [Prop. 1.47]. nd the (sum of the squares) on ὅταναἱἀπὸτοῦκέντρουἐπ αὐτὰςκάθετοιἀγόμεναιἴσαι and (is) equal to the (square) on. or the angle ὦσιν αἱἄρα,δἴσονἀπέχουσινἀπὸτοῦκέντρου. at (is) a right-angle [Prop. 1.47]. Thus, the (sum of Ἀλλὰδὴαἱ,Δεὐθεῖαιἴσονἀπεχέτωσανἀπὸτοῦ the squares) on and is equal to the (sum of the κέντρου,τουτέστινἴσηἔστωἡτῇ.λέγω,ὅτιἴση squares) on and, of which the (square) on ἐστὶκαὶἡτῇδ. is equal to the (square) on. or is equal to. 84
Τῶν γὰρ αὐτῶν κατασκευασθέντων ὁμοίως δείξομεν, Thus, the remaining (square) on is equal to the (re- ὅτιδιπλῆἐστινἡμὲντῆς,ἡδὲδτῆς καὶἐπεὶ maining square) on. Thus, (is) equal to. nd ἴσηἐστὶνἡτῇ,ἴσονἐστὶτὸἀπὸτῆςτῷἀπὸ straight-lines in a circle are said to be equally far from τῆς ἀλλὰ τῷ μὲν ἀπὸ τῆς ἴσα ἐστὶ τὰ ἀπὸ τῶν, the center when perpendicular (straight-lines) which are,τῷδὲἀπὸτῆςἴσατὰἀπὸτῶν,.τὰἄραἀπὸ drawn to them from the center are equal [ef. 3.4]. Thus, τῶν, ἴσα ἐστὶ τοῖς ἀπὸ τῶν, ὧν τὸ ἀπὸ τῆς and are equally far from the center. τῷἀπὸτῆςἐστινἴσον ἴσηγὰρἡτῇ λοιπὸν So, let the straight-lines and be equally far ἄρατὸἀπὸτῆςἴσονἐστὶτῷἀπὸτῆς ἴσηἄραἡ from the center. That is to say, let be equal to. I τῇ καί ἐστι τῆς μὲν διπλῆ ἡ, τῆς δὲ διπλῆ say that is also equal to. ἡδ ἴσηἄραἡτῇδ. or, with the same construction, we can, similarly, ν κύκλῳ ἄρα αἱ ἴσαι εὐθεῖαι ἴσον ἀπέχουσιν ἀπὸ show that is double, and (double). nd τοῦκέντρου,καὶαἱἴσονἀπέχουσαιἀπὸτοῦκέντρουἴσαι since is equal to, the (square) on is equal to ἀλλήλαις εἰσίν ὅπερ ἔδει δεῖξαι. the (square) on. ut, the (sum of the squares) on and is equal to the (square) on [Prop. 1.47]. nd the (sum of the squares) on and (is) equal to the (square) on [Prop. 1.47]. Thus, the (sum of the squares) on and is equal to the (sum of the squares) on and, of which the (square) on is equal to the (square) on. or (is) equal to. Thus, the remaining (square) on is equal to the (remaining square) on. Thus, (is) equal to. nd is double, and double. Thus, (is) equal to. Thus, in a circle, equal straight-lines are equally far from the center, and (straight-lines) which are equally far from the center are equal to one another. (Which is) the very thing it was required to show. The reek text has, which is obviously a mistake.. Proposition 15 ν κύκλῳ μεγίστη μὲν ἡ διάμετρος, τῶν δὲ ἄλλων ἀεὶ In a circle, a diameter (is) the greatest (straight-line), ἡ ἔγγιον τοῦ κέντρου τῆς ἀπώτερον μείζων ἐστίν. and for the others, a (straight-line) nearer to the center στω κύκλος ὁ Δ, διάμετρος δὲ αὐτοῦ ἔστω ἡ Δ, is always greater than one further away. κέντρονδὲτὸ,καὶἔγγιονμὲντῆςδδιαμέτρουἔστωἡ Let be a circle, and let be its diameter,, ἀπώτερον δὲ ἡ λέγω, ὅτι μεγίστη μέν ἐστιν ἡ Δ, and (its) center. nd let be nearer to the diameter μείζωνδὲἡτῆς., and further away. I say that is the greatest χθωσαν γὰρ ἀπὸ τοῦ κέντρου ἐπὶ τὰς, (straight-line), and (is) greater than. κάθετοιαἱθ,κ.καὶἐπεὶἔγγιονμὲντοῦκέντρουἐστὶν or let H and K have been drawn from the cen- ἡ, ἀπώτερον δὲ ἡ, μείζων ἄρα ἡ Κ τῆς Θ. κείσθω ter, at right-angles to and (respectively) τῇθἴσηἡλ,καὶδιὰτοῦλτῇκπρὸςὀρθὰςἀχθεῖσα [Prop. 1.12]. nd since is nearer to the center, ἡλμδιήχθωἐπὶτὸν,καὶἐπεζεύχθωσαναἱμ,ν,, and further away, K (is) thus greater than H. [ef. 3.5]. Let L be made equal to H [Prop. 1.3]. ΚαὶἐπεὶἴσηἐστὶνἡΘτῇΛ,ἴσηἐστὶκαὶἡτῇ nd LM being drawn through L, at right-angles to K ΜΝ.πάλιν,ἐπεὶἴσηἐστὶνἡμὲντῇΜ,ἡδὲΔτῇ [Prop. 1.11], let it have been drawn through to N. nd Ν,ἡἄραΔταῖςΜ,Νἴσηἐστίν.ἀλλ αἱμὲνμ,ν let M, N,, and have been joined. τῆςμνμείζονέςεἰσιν[καὶἡδτῆςμνμείζωνἐστίν], nd since H is equal to L, is also equal to ἴσηδὲἡμντῇ ἡδἄρατῆςμείζωνἐστίν. καὶ MN [Prop. 3.14]. gain, since is equal to M, and ἐπεὶδύοαἱμ,νδύοταῖς,ἴσαιεἰσίν,καὶγωνία to N, is thus equal to M and N. ut, M ἡὑπὸμνγωνίαςτῆςὑπὸμείζων[ἐστίν],βάσιςἄρα and N is greater than MN [Prop. 1.20] [also is 85
ἡμνβάσεωςτῆςμείζωνἐστίν. ἀλλὰἡμντῇ greater than MN], and MN (is) equal to. Thus, ἐδείχθη ἴση[καὶ ἡ τῆς μείζων ἐστίν]. μεγίστη μὲν is greater than. nd since the two (straight-lines) ἄραἡδδιάμετρος,μείζωνδὲἡτῆς. M, N are equal to the two (straight-lines), (respectively), and angle M N [is] greater than angle, the base MN is thus greater than the base [Prop. 1.24]. ut, MN was shown (to be) equal to [(so) is also greater than ]. Thus, the diameter (is) the greatest (straight-line), and (is) greater than. Μ M Κ Λ Θ K L H Ν N ν κύκλῳ ἄρα μεγίστη μὲν έστιν ἡ διάμετρος, τῶν δὲ Thus, in a circle, a diameter (is) the greatest (straight- ἄλλων ἀεὶ ἡ ἔγγιον τοῦ κέντρου τῆς ἀπώτερον μείζων ἐστίν line), and for the others, a (straight-line) nearer to the ὅπερ ἔδει δεῖξαι. center is always greater than one further away. (Which is) the very thing it was required to show. uclid should have said to the center, rather than to the diameter, since, and are not necessarily parallel. This is not proved, except by reference to the figure.. Proposition 16 τῇ διαμέτρῳ τοῦ κύκλου πρὸς ὀρθὰς ἀπ ἄκρας (straight-line) drawn at right-angles to the diameter ἀγομένηἐκτὸςπεσεῖταιτοῦκύκλου, καὶεἰςτὸν μεταξὺ of a circle, from its end, will fall outside the circle. nd τόπον τῆς τε εὐθείας καὶ τῆς περιφερείας ἑτέρα εὐθεῖα οὐ another straight-line cannot be inserted into the space beπαρεμπεσεῖται, καὶ ἡ μὲν τοῦ ἡμικυκλίου γωνία ἁπάσης tween the (aforementioned) straight-line and the circumγωνίας ὀξείας εὐθυγράμμου μείζων ἐστίν, ἡ δὲ λοιπὴ ference. nd the angle of the semi-circle is greater than ἐλάττων. any acute rectilinear angle whatsoever, and the remain- στω κύκλος ὁ περὶ κέντρον τὸ Δ καὶ διάμετρον ing (angle is) less (than any acute rectilinear angle). τὴν λέγω,ὅτιἡἀπὸτοῦτῇπρὸςὀρθὰςἀπ Let be a circle around the center and the di- ἄκρας ἀγομένη ἐκτὸς πεσεῖται τοῦ κύκλου. ameter. I say that the (straight-line) drawn from, Μὴγάρ,ἀλλ εἰδυνατόν,πιπτέτωἐντὸςὡςἡ,καὶ at right-angles to [Prop 1.11], from its end, will fall ἐπεζεύχθωἡδ. outside the circle. πεὶἴσηἐστὶνἡδτῇδ,ἴσηἐστὶκαὶγωνίαἡὑπὸ or (if) not then, if possible, let it fall inside, like Δ γωνίᾳ τῇ ὑπὸ Δ. ὀρθὴ δὲ ἡ ὑπὸ Δ ὀρθὴ ἄρα (in the figure), and let have been joined. καὶἡὑπὸδ τριγώνουδὴτοῦδαἱδύογωνίαιαἱ Since is equal to, angle is also equal ὑπὸ Δ, Δ δύο ὀρθαῖς ἴσαι εἰσίν ὅπερ ἐστὶν ἀδύνατον. to angle [Prop. 1.5]. nd (is) a right-angle. οὐκἄραἡἀπὸτοῦσημείουτῇπρὸςὀρθὰςἀγομένη Thus, (is) also a right-angle. So, in triangle, ἐντὸςπεσεῖταιτοῦκύκλου. ὁμοίωςδὴδεῖξομεν,ὅτιοὐδ the two angles and are equal to two right- ἐπὶ τῆς περιφερείας ἐκτὸς ἄρα. angles. The very thing is impossible [Prop. 1.17]. Thus, the (straight-line) drawn from point, at right-angles 86
to, will not fall inside the circle. So, similarly, we can show that neither (will it fall) on the circumference. Thus, (it will fall) outside (the circle). Θ Πιπτέτωὡςἡ λέγωδή,ὅτιεἰςτὸνμεταξὺτόπον Let it fall like (in the figure). So, I say that another τῆς τε εὐθείας καὶ τῆς Θ περιφερείας ἑτέρα εὐθεῖα straight-line cannot be inserted into the space between οὐπαρεμπεσεῖται. the straight-line and the circumference H. ἰγὰρδυνατόν,παρεμπιπτέτωὡςἡ,καὶἤχθωἀπὸ or, if possible, let it be inserted like (in the figτοῦδσημείουἐπὶτῆνκάθετοςἡδ.καὶἐπεὶὀρθή ure), and let have been drawn from point, perpen- ἐστινἡὑπὸδ,ἐλάττωνδὲὀρθῆςἡὑπὸδ,μείζων dicular to [Prop. 1.12]. nd since is a right- ἄραἡδτῆςδ.ἴσηδὲἡδτῇδθ μείζωνἄραἡδθ angle, and (is) less than a right-angle, (is) τῆς Δ, ἡ ἐλάττων τῆς μείζονος ὅπερ ἐστὶν ἀδύνατον. οὐκ thus greater than [Prop. 1.19]. nd (is) equal ἄραεἰςτὸνμεταξὺτόποντῆςτεεὐθείαςκαὶτῆςπεριφερείας to H. Thus, H (is) greater than, the lesser than ἑτέρα εὐθεῖα παρεμπεσεῖται. the greater. The very thing is impossible. Thus, another Λέγω, ὅτι καὶ ἡ μὲν τοῦ ἡμικυκλίου γωνία ἡ περιεχομένη straight-line cannot be inserted into the space between ὑπό τε τῆς εὐθείας καὶ τῆς Θ περιφερείας ἁπάσης the straight-line () and the circumference. γωνίας ὀξείας εὐθυγράμμου μείζων ἐστίν, ἡ δὲ λοιπὴ ἡ πε- nd I also say that the semi-circular angle contained ριεχομένη ὑπό τε τῆς Θ περιφερείας καὶ τῆς εὐθείας by the straight-line and the circumference H is ἁπάσης γωνίας ὀξείας εὐθυγράμμου ἐλάττων ἐστίν. greater than any acute rectilinear angle whatsoever, and ἰ γὰρ ἐστί τις γωνία εὐθύγραμμος μείζων μὲν τῆς the remaining (angle) contained by the circumference περιεχομένης ὑπό τε τῆς εὐθείας καὶ τῆς Θ περι- H and the straight-line is less than any acute recφερείας, ἐλάττων δὲ τῆς περιεχομένης ὑπό τε τῆς Θ tilinear angle whatsoever. περιφερείας καὶ τὴς εὐθείας, εἰς τὸν μεταξὺ τόπον τῆς or if any rectilinear angle is greater than the (anτε Θ περιφερείας καὶ τῆς εὐθείας εὐθεῖα παρεμ- gle) contained by the straight-line and the circumπεσεῖται, ἥτις ποιήσει μείζονα μὲν τῆς περιεχομένης ὑπὸ ference H, or less than the (angle) contained by the τε τῆς εὐθείας καὶ τῆς Θ περιφερείας ὑπὸ εὐθειῶν circumference H and the straight-line, then a περιεχομένην, ἐλάττονα δὲ τῆς περιεχομένης ὑπό τε τῆς straight-line can be inserted into the space between the Θ περιφερείας καὶ τῆς εὐθείας. οὐ παρεμπίπτει δέ circumference H and the straight-line anything οὐκ ἄρα τῆς περιεχομένης γωνίας ὑπό τε τῆς εὐθείας which will make (an angle) contained by straight-lines καὶ τῆς Θ περιφερείας ἔσται μείζων ὀξεῖα ὑπὸ εὐθειῶν greater than the angle contained by the straight-line περιεχομένη, οὐδὲ μὴν ἐλάττων τῆς περιεχομένης ὑπό τε and the circumference H, or less than the (angle) τῆς Θ περιφερείας καὶ τῆς εὐθείας. contained by the circumference H and the straightline. ut (such a straight-line) cannot be inserted. Thus, an acute (angle) contained by straight-lines cannot be greater than the angle contained by the straight-line and the circumference H, neither (can it be) less than the (angle) contained by the circumference H and the straight-line. H 87