§1.prop.16παντὸς τριγώνου μιᾶς τῶν πλευρῶν προσεκβληθείσης ἡ ἐκτὸς γωνία ἑκατέρας τῶν ἐντὸς καὶ ἀπεναντίον γωνιῶν μείζων ἐστίν.
In any triangle, if one of the sides is produced, the exterior angle is greater than either of the interior and opposite angles.
ἔστω τρίγωνον τὸ ΑΒΓ, καὶ προσεκβεβλήσθω αὐτοῦ μία πλευρὰ ἡ ΒΓ ἐπὶ τὸ Δ· λέγω, ὅτι ἡ ἐκτὸς γωνία ἡ ὑπὸ ΑΓΔ μείζων ἐστὶν ἑκατέρας τῶν ἐντὸς καὶ ἀπεναντίον τῶν ὑπὸ ΓΒΑ, ΒΑΓ γωνιῶν.
Let ABC be a triangle, and let one side of it, BC, be produced to D; I say that the exterior angle ACD is greater than either of the interior and opposite angles CBA, BAC.
τετμήσθω ἡ ΑΓ δίχα κατὰ τὸ Ε, καὶ ἐπιζευχθεῖσα ἡ ΒΕ ἐκβεβλήσθω ἐπʼ εὐθείας ἐπὶ τὸ Ζ, καὶ κείσθω τῇ ΒΕ ἴση ἡ ΕΖ, καὶ ἐπεζεύχθω ἡ ΖΓ, καὶ διήχθω ἡ ΑΓ ἐπὶ τὸ Η.
ἐπεὶ οὖν ἴση ἐστὶν ἡ μὲν ΑΕ τῇ ΕΓ, ἡ δὲ ΒΕ τῇ ΕΖ, δύο δὴ αἱ ΑΕ, ΕΒ δυσὶ ταῖς ΓΕ, ΕΖ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ·
Let AC be bisected at E, and let BE, being joined, be produced in a straight line to Z, and let EZ be made equal to BE, and let ZC be joined, and let AC be drawn through to H.
καὶ γωνία ἡ ὑπὸ ΑΕΒ γωνίᾳ τῇ ὑπὸ ΖΕΓ ἴση ἐστίν· κατὰ κορυφὴν γάρ· βάσις ἄρα ἡ ΑΒ βάσει τῇ ΖΓ ἴση ἐστίν, καὶ τὸ ΑΒΕ τρίγωνον τῷ ΖΕΓ τριγώνῳ ἐστὶν ἴσον, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ, ὑφʼ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν·
Since, therefore, AE is equal to EC, and BE to EZ, the two sides AE, EB are therefore equal to the two sides GE, EZ respectively; and the angle AEB is equal to the angle ZEG; for they are vertical; therefore the base AB is equal to the base ZC, and the triangle ABE is equal to the triangle ZEG, and the remaining angles are equal to the remaining angles respectively, namely those which the equal sides subtend; therefore the angle BAE is equal to the angle EGZ.
ἴση ἄρα ἐστὶν ἡ ὑπὸ ΒΑΕ τῇ ὑπὸ ΕΓΖ. μείζων δέ ἐστιν ἡ ὑπὸ ΕΓΔ τῆς ὑπὸ ΕΓΖ·
But the angle EGD is greater than the angle EGZ; therefore the angle ACD is greater than the angle BAE.
μείζων ἄρα ἡ ὑπὸ ΑΓΔ τῆς ὑπὸ ΒΑΕ. ὁμοίως δὴ τῆς ΒΓ τετμημένης δίχα δειχθήσεται καὶ ἡ ὑπὸ ΒΓΗ, τουτέστιν ἡ ὑπὸ ΑΓΔ, μείζων καὶ τῆς ὑπὸ ΑΒΓ.
παντὸς ἄρα τριγώνου μιᾶς τῶν πλευρῶν προσεκβληθείσης ἡ ἐκτὸς γωνία ἑκατέρας τῶν ἐντὸς καὶ ἀπεναντίον γωνιῶν μείζων ἐστίν·
Similarly, if BC is bisected, the angle BCH, that is, the angle ACD, can be proved to be greater than the angle ABC as well.
ὅπερ ἔδει δεῖξαι.
Therefore, in any triangle, if one of the sides is produced, the exterior angle is greater than either of the interior and opposite angles. - Being what it was required to prove.
§1.prop.17παντὸς τριγώνου αἱ δύο γωνίαι δύο ὀρθῶν ἐλάσσονές εἰσι πάντῃ μεταλαμβανόμεναι.
In any triangle, two angles taken together in any manner are less than two right angles.
ἔστω τρίγωνον τὸ ΑΒΓ· λέγω, ὅτι τοῦ ΑΒΓ τριγώνου αἱ δύο γωνίαι δύο ὀρθῶν ἐλάττονές εἰσι πάντῃ μεταλαμβανόμεναι.
Let ABC be a triangle; I say that two angles of the triangle ABC taken together in any manner are less than two right angles.
Ἐκβεβλήσθω γὰρ ἡ ΒΓ ἐπὶ τὸ Δ.
καὶ ἐπεὶ τριγώνου τοῦ ΑΒΓ ἐκτός ἐστι γωνία ἡ ὑπὸ ΑΓΔ, μείζων ἐστὶ τῆς ἐντὸς καὶ ἀπεναντίον τῆς ὑπὸ ΑΒΓ. κοινὴ προσκείσθω ἡ ὑπὸ ΑΓΒ· αἱ ἄρα ὑπὸ ΑΓΔ, ΑΓΒ τῶν ὑπὸ ΑΒΓ, ΒΓΑ μείζονές εἰσιν.
For let BC be produced to D. And since, of the triangle ABC, the angle ACD is an exterior angle, it is greater than the interior and opposite angle ABC. Let the common angle ACB be added; therefore the angles ACD, ACB are greater than the angles ABC, BCA.
ἀλλʼ αἱ ὑπὸ ΑΓΔ, ΑΓΒ δύο ὀρθαῖς ἴσαι εἰσίν· αἱ ἄρα ὑπὸ ΑΒΓ, ΒΓΑ δύο ὀρθῶν ἐλάσσονές εἰσιν.
But the angles ACD, ACB are equal to two right angles; therefore the angles ABC, BCA are less than two right angles.
ὁμοίως δὴ δείξομεν, ὅτι καὶ αἱ ὑπὸ ΒΑΓ, ΑΓΒ δύο ὀρθῶν ἐλάσσονές εἰσι καὶ ἔτι αἱ ὑπὸ ΓΑΒ, ΑΒΓ.
παντὸς ἄρα τριγώνου αἱ δύο γωνίαι δύο ὀρθῶν ἐλάσσονές εἰσι πάντῃ μεταλαμβανόμεναι·
Similarly, we can prove that the angles BAC, ACB are also less than two right angles, and further the angles CAB, ABC as well.
ὅπερ ἔδει δεῖξαι.
Therefore, in any triangle, two angles taken together in any manner are less than two right angles. - Being what it was required to prove.
§1.prop.18παντὸς τριγώνου ἡ μείζων πλευρὰ τὴν μείζονα γωνίαν ὑποτείνει.
In any triangle, the greater side subtends the greater angle.
ἔστω γὰρ τρίγωνον τὸ ΑΒΓ μείζονα ἔχον τὴν ΑΓ πλευρὰν τῆς ΑΒ· λέγω, ὅτι καὶ γωνία ἡ ὑπὸ ΑΒΓ μείζων ἐστὶ τῆς ὑπὸ ΒΓΑ.
ἐπεὶ γὰρ μείζων ἐστὶν ἡ ΑΓ τῆς ΑΒ, κείσθω τῇ ΑΒ ἴση ἡ ΑΔ, καὶ ἐπεζεύχθω ἡ ΒΔ.
καὶ ἐπεὶ τριγώνου τοῦ ΒΓΔ ἐκτός ἐστι γωνία ἡ ὑπὸ ΑΔΒ, μείζων ἐστὶ τῆς ἐντὸς καὶ ἀπεναντίον τῆς ὑπὸ ΔΓΒ·
For let ABC be a triangle having the side AC greater than AB; I say that the angle ABC is also greater than the angle BCA. For since AC is greater than AB, let AD be made equal to AB, and let BD be joined. And since, of the triangle BCD, the angle ADB is an exterior angle, it is greater than the interior and opposite angle DCB.
ἴση δὲ ἡ ὑπὸ ΑΔΒ τῇ ὑπὸ ΑΒΔ, ἐπεὶ καὶ πλευρὰ ἡ ΑΒ τῇ ΑΔ ἐστιν ἴση· μείζων ἄρα καὶ ἡ ὑπὸ ΑΒΔ τῆς ὑπὸ ΑΓΒ· πολλῷ ἄρα ἡ ὑπὸ ΑΒΓ μείζων ἐστὶ τῆς ὑπὸ ΑΓΒ.
παντὸς ἄρα τριγώνου ἡ μείζων πλευρὰ τὴν μείζονα γωνίαν ὑποτείνει·
But the angle ADB is equal to the angle ABD, since the side AB is also equal to AD; therefore the angle ABD is also greater than the angle ACB; therefore the angle ABC is much greater than the angle ACB.
ὅπερ ἔδει δεῖξαι.
Therefore, in any triangle, the greater side subtends the greater angle. - Being what it was required to prove.