§11.prop.22ἐὰν ὦσι τρεῖς γωνίαι ἐπίπεδοι, ὧν αἱ δύο τῆς λοιπῆς μείζονές εἰσι πάντῃ μεταλαμβανόμεναι, περιέχωσι δὲ αὐτὰς ἴσαι εὐθεῖαι, δυνατόν ἐστιν ἐκ τῶν ἐπιζευγνυουσῶν τὰς ἴσας εὐθείας τρίγωνον συστήσασθαι.
If there be three plane angles, of which two, taken together in any way, are greater than the remaining one, and they are contained by equal straight lines, it is possible to construct a triangle out of the straight lines joining the equal straight lines.
ἔστωσαν τρεῖς γωνίαι ἐπίπεδοι αἱ ὑπὸ ΑΒΓ, ΔΕΖ, ΗΘΚ, ὧν αἱ δύο τῆς λοιπῆς μείζονές εἰσι πάντῃ μεταλαμβανόμεναι, αἱ μὲν ὑπὸ ΑΒΓ, ΔΕΖ τῆς ὑπὸ ΗΘΚ, αἱ δὲ ὑπὸ ΔΕΖ, ΗΘΚ τῆς ὑπὸ ΑΒΓ, καὶ ἔτι αἱ ὑπὸ ΗΘΚ, ΑΒΓ τῆς ὑπὸ ΔΕΖ, καὶ ἔστωσαν ἴσαι αἱ ΑΒ, ΒΓ, ΔΕ, ΕΖ, ΗΘ, ΘΚ εὐθεῖαι, καὶ ἐπεζεύχθωσαν αἱ ΑΓ, ΔΖ, ΗΚ· λέγω, ὅτι δυνατόν ἐστιν ἐκ τῶν ἴσων ταῖς ΑΓ, ΔΖ, ΗΚ τρίγωνον συστήσασθαι, τουτέστιν ὅτι τῶν ΑΓ, ΔΖ, ΗΚ δύο ὁποιαιοῦν τῆς λοιπῆς μείζονές εἰσιν.
Let there be three plane angles ABC, DEF, GHK, of which two, taken together in any way, are greater than the remaining one, namely the angles ABC, DEF greater than GHK, the angles DEF, GHK greater than ABC, and further the angles GHK, ABC greater than DEF; and let the straight lines AB, BC, DE, EF, GH, HK be equal, and let AC, DF, GK be joined; I say that it is possible to construct a triangle out of straight lines equal to AC, DF, GK, that is, that any two of AC, DF, GK are greater than the remaining one.
εἰ μὲν οὖν αἱ ὑπὸ ΑΒΓ, ΔΕΖ, ΗΘΚ γωνίαι ἴσαι ἀλλήλαις εἰσίν, φανερόν, ὅτι καὶ τῶν ΑΓ, ΔΖ, ΗΚ ἴσων γινομένων δυνατόν ἐστιν ἐκ τῶν ἴσων ταῖς ΑΓ, ΔΖ, ΗΚ τρίγωνον συστήσασθαι.
If then the angles ABC, DEF, GHK are equal to one another, it is manifest that, AC, DF, GK also becoming equal, it is possible to construct a triangle out of straight lines equal to AC, DF, GK.
εἰ δὲ οὔ, ἔστωσαν ἄνισοι, καὶ συνεστάτω πρὸς τῇ ΘΚ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Θ τῇ ὑπὸ ΑΒΓ γωνίᾳ ἴση ἡ ὑπὸ ΚΘΛ· καὶ κείσθω μιᾷ τῶν ΑΒ, ΒΓ, ΔΕ, ΕΖ, ΗΘ, ΘΚ ἴση ἡ ΘΛ, καὶ ἐπεζεύχθωσαν αἱ ΚΛ, ΗΛ. καὶ ἐπεὶ δύο αἱ ΑΒ, ΒΓ δυσὶ ταῖς ΚΘ, ΘΛ ἴσαι εἰσίν, καὶ γωνία ἡ πρὸς τῷ Β γωνίᾳ τῇ ὑπὸ ΚΘΛ ἴση, βάσις ἄρα ἡ ΑΓ βάσει τῇ ΚΛ ἴση.
But if not, let them be unequal, and let there be constructed on the straight line HK and at the point H on it, the angle KHL equal to the angle ABC; and let HL be made equal to one of AB, BC, DE, EF, GH, HK, and let KL, GL be joined. And since the two sides AB, BC are equal to the two sides KH, HL, and the angle at B is equal to the angle KHL, therefore the base AC is equal to the base KL.
καὶ ἐπεὶ αἱ ὑπὸ ΑΒΓ, ΗΘΚ τῆς ὑπὸ ΔΕΖ μείζονές εἰσιν, ἴση δὲ ἡ ὑπὸ ΑΒΓ τῇ ὑπὸ ΚΘΛ, ἡ ἄρα ὑπὸ ΗΘΛ τῆς ὑπὸ ΔΕΖ μείζων ἐστίν.
And since the angles ABC, GHK are greater than DEF, and ABC is equal to KHL, therefore the angle GHL is greater than DEF.
καὶ ἐπεὶ δύο αἱ ΗΘ, ΘΛ δύο ταῖς ΔΕ, ΕΖ ἴσαι εἰσίν, καὶ γωνία ἡ ὑπὸ ΗΘΛ γωνίας τῆς ὑπὸ ΔΕΖ μείζων, βάσις ἄρα ἡ ΗΛ βάσεως τῆς ΔΖ μείζων ἐστίν.
And since the two sides GH, HL are equal to the two sides DE, EF, and the angle GHL is greater than the angle DEF, therefore the base GL is greater than the base DF.
ἀλλὰ αἱ ΗΚ, ΚΛ τῆς ΗΛ μείζονές εἰσιν.
But GK, KL are greater than GL.
πολλῷ ἄρα αἱ ΗΚ, ΚΛ τῆς ΔΖ μείζονές εἰσιν.
Therefore GK, KL are much greater than DF.
ἴση δὲ ἡ ΚΛ τῇ ΑΓ· αἱ ΑΓ, ΗΚ ἄρα τῆς λοιπῆς τῆς ΔΖ μείζονές εἰσιν.
And KL is equal to AC; therefore AC, GK are greater than the remaining DF.
ὁμοίως δὴ δείξομεν, ὅτι καὶ αἱ μὲν ΑΓ, ΔΖ τῆς ΗΚ μείζονές εἰσιν, καὶ ἔτι αἱ ΔΖ, ΗΚ τῆς ΑΓ μείζονές εἰσιν.
Similarly indeed we shall show that AC, DF are also greater than GK, and further DF, GK are greater than AC.
δυνατὸν ἄρα ἐστὶν ἐκ τῶν ἴσων ταῖς ΑΓ, ΔΖ, ΗΚ τρίγωνον συστήσασθαι· ὅπερ ἔδει δεῖξαι.
Therefore it is possible to construct a triangle out of straight lines equal to AC, DF, GK; which was to be proved.