Constructions for inscribed and circumscribed figures: Book 4 Proposition 5
Translations
Περὶ τὸ δοθὲν τρίγωνον κύκλον περιγράψαι. Ἔστω τὸ δοθὲν τρίγωνον τὸ ΑΒΓ: δεῖ <δὴ>περὶ τὸ δοθὲν τρίγωνον τὸ ΑΒΓ κύκλον περιγράψαι. Τετμήσθωσαν αἱ ΑΒ, ΑΓ εὐθεῖαι δίχα κατὰ τὰ Δ, Ε σημεῖα, καὶ ἀπὸ τῶν Δ, Ε σημείων ταῖς ΑΒ, ΑΓ πρὸς ὁρθὰς ἤχθωσαν αἱ ΔΖ, ΕΖ: συμπεσοῦνται δὴ ἤτοι ἐντὸς τοῦ ΑΒΓ τριγώνου ἢ ἐπὶ τῆς ΒΓ εὐθείας ἢ ἐκτὸς τῆς ΒΓ. Συμπιπτέτωσαν πρότερον ἐντὸς κατὰ τὸ Ζ, καὶ ἐπεζεύχθωσαν αἱ ΖΒ, ΖΓ, ΖΑ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΔ τῇ ΔΒ, κοινὴ δὲ καὶ πρὸς ὀρθὰς ἡ ΔΖ, βάσις ἄρα ἡ ΑΖ βάσει τῇ ΖΒ ἐστιν ἴση. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἡ ΓΖ τῇ ΑΖ ἐστιν ἴση: ὥστε καὶ ἡ ΖΒ τῇ ΖΓ ἐστιν ἴση: αἱ τρεῖς ἄρα αἱ ΖΑ, ΖΒ, ΖΓ ἴσαι ἀλλήλαις εἰσίν. ὁ ἄρα κέντρῳ τῷ Ζ διαστήματι δὲ ἑνὶ τῶν Α, Β, Γ κύκλος γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν σημείων, καὶ ἔσται περιγεγραμμένος ὁ κύκλος περὶ τὸ ΑΒΓ τρίγωνον. περιγεγράφθω ὡς ὁ ΑΒΓ. Ἀλλὰ δὴ αἱ ΔΖ, ΕΖ συμπιπτέτωσαν ἐπὶ τῆς ΒΓ εὐθείας κατὰ τὸ Ζ, ὡς ἔχει ἐπὶ τῆς δευτέρας καταγραφῆς, καὶ ἐπεζεύχθω ἡ ΑΖ. ὁμοίως δὴ δείξομεν, ὅτι τὸ Ζ σημεῖον κέντρον ἐστὶ τοῦ περὶ τὸ ΑΒΓ τρίγωνον περιγραφομένου κύκλου. Ἀλλὰ δὴ αἱ ΔΖ, ΕΖ συμπιπτέτωσαν ἐκτὸς τοῦ ΑΒΓ τριγώνου κατὰ τὸ Ζ πάλιν, ὡς ἔχει ἐπὶ τῆς τρίτης καταγραφῆς, καὶ ἐπεζεύχθωσαν αἱ ΑΖ, ΒΖ, ΓΖ. καὶ ἐπεὶ πάλιν ἴση ἐστὶν ἡ ΑΔ τῇ ΔΒ, κοινὴ δὲ καὶ πρὸς ὀρθὰς ἡ ΔΖ, βάσις ἄρα ἡ ΑΖ βάσει τῇ ΒΖ ἐστιν ἴση. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἡ ΓΖ τῇ ΑΖ ἐστιν ἴση: ὥστε καὶ ἡ ΒΖ τῇ ΖΓ ἐστιν ἴση: ὁ ἄρα [πάλιν] κέντρῳ τῷ Ζ διαστήματι δὲ ἑνὶ τῶν ΖΑ, ΖΒ, ΖΓ κύκλος γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν σημείων, καὶ ἔσται περιγεγραμμένος περὶ τὸ ΑΒΓ τρίγωνον. Περὶ τὸ δοθὲν ἄρα τρίγωνον κύκλος περιγέγραπται: ὅπερ ἔδει ποιῆσαι.[Πόρισμα.]Καὶ φανερόν, ὅτι, ὅτε μὲν ἐντὸς τοῦ τριγώνου πίπτει τὸ κέντρον τοῦ κύκλου, ἡ ὑπὸ ΒΑΓ γωνία ἐν μείζονι τμήματι τοῦ ἡμικυκλίου τυγχάνουσα ἐλάττων ἐστὶν ὀρθῆς: ὅτε δὲ ἐπὶ τῆς ΒΓ εὐθείας τὸ κέντρον πίπτει, ἡ ὑπὸ ΒΑΓ γωνία ἐν ἡμικυκλίῳ τυγχάνουσα ὀρθή ἐστιν: ὅτε δὲ τὸ κέντρον τοῦ κύκλου ἐκτὸς τοῦ τριγώνου πίπτει, ἡ ὑπὸ ΒΑΓ ἐν ἐλάττονι τμήματι τοῦ ἡμικυκλίου τυγχάνουσα μείζων ἐστὶν ὀρθῆς. [ὥστε καὶ ὅταν ἐλάττων ὀρθῆς τυγχάνῃ ἡ διδομένη γωνία, ἐντὸς τοῦ τριγώνου πεσοῦνται αἱ ΔΖ, ΕΖ, ὅταν δὲ ὀρθή, ἐπὶ τῆς ΒΓ, ὅταν δὲ μείζων ὀρθῆς, ἐκτὸς τῆς ΒΓ: ὅπερ ἔδει ποιῆσαι.]
About a given triangle to circumscribe a circle. Let ABC be the given triangle; thus it is required to circumscribe a circle about the given triangle ABC. Let the straight lines AB, AC be bisected at the points D, E [I. 10], and from the points D, E let DF, EF be drawn at right angles to AB, AC; they will then meet within the triangle ABC, or on the straight line BC, or outside BC. First let them meet within at F, and let FB, FC, FA be joined. Then, since AD is equal to DB, and DF is common and at right angles, therefore the base AF is equal to the base FB. [I. 4] Similarly we can prove that CF is also equal to AF; so that FB is also equal to FC; therefore the three straight lines FA, FB, FC are equal to one another. Therefore the circle described with centre F and distance one of the straight lines FA, FB, FC will pass also through the remaining points, and the circle will have been circumscribed about the triangle ABC. Let it be circumscribed, as ABC. Next, let DF, EF meet on the straight line BC at F, as is the case in the second figure; and let AF be joined. Then, similarly, we shall prove that the point F is the centre of the circle circumscribed about the triangle ABC. Again, let DF, EF meet outside the triangle ABC at F, as is the case in the third figure, and let AF, BF, CF be joined. Then again, since AD is equal to DB, and DF is common and at right angles, therefore the base AF is equal to the base BF. [I. 4] Similarly we can prove that CF is also equal to AF; so that BF is also equal to FC; therefore the circle described with centre F and distance one of the straight lines FA, FB, FC will pass also through the remaining points, and will have been circumscribed about the triangle ABC. Therefore about the given triangle a circle has been circumscribed. Q. E. F.[Porism.] And it is manifest that, when the centre of the circle falls within the triangle, the angle BAC, being in a segment greater than the semicircle, is less than a right angle; when the centre falls on the straight line BC, the angle BAC, being in a semicircle, is right; and when the centre of the circle falls outside the triangle, the angle BAC, being in a segment less than the semicircle, is greater than a right angle. [III. 31]. [So that, further, when the given angle happens to be less than a right angle, DF, EF will fall within the triangle, when it is right, on BC, and, when it is greater than a right angle, outside BC: (being) what it was required to do.]