§4.prop.5περὶ τὸ δοθὲν τρίγωνον κύκλον περιγράψαι.
About the given triangle to circumscribe a circle.
ἔστω τὸ δοθὲν τρίγωνον τὸ ΑΒΓ· δεῖ 2 περὶ τὸ δοθὲν τρίγωνον τὸ ΑΒΓ κύκλον περιγράψαι.
Let ABC be the given triangle; it is required then about the given triangle ABC to circumscribe a circle.
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·
Let the straight lines AB, AC be bisected at the points D, E, and from the points D, E let DZ, EZ be drawn at right angles to AB, AC; they will then meet either within the triangle ABC, or on the straight line BC, or outside BC.
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.
Let them first meet within at Z, and let ZB, ZC, ZA be joined. And since AD is equal to DB, and DZ is common and at right angles, therefore the base AZ is equal to the base ZB.
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.
Similarly then we will prove that CZ is also equal to AZ; so that ZB is also equal to ZC; therefore the three straight lines ZA, ZB, ZC are equal to one another.
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.
Therefore the circle described with center Z and distance one of the points A, B, C will pass also through the remaining points, and the circle will have been circumscribed about the triangle ABC.
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 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. 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.
Let it be circumscribed as ABC. But now let DZ, EZ meet on the straight line BC at Z, as in the second figure, and let AZ be joined....