Euclid · Elements §1.prop.26

Congruence of Triangles with Two Angles and One Side Equal

Passage 15 of 316 · Greek

Summary

Proves, using proof by contradiction, that two triangles are congruent if they have two angles and one side (either the included side or a side subtending one of the angles) equal respectively.

§1.prop.26ἐὰν δύο τρίγωνα τὰς δύο γωνίας δυσὶ γωνίαις ἴσας ἔχῃ ἑκατέραν ἑκατέρᾳ καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην ἤτοι τὴν πρὸς ταῖς ἴσαις γωνίαις ἢ τὴν ὑποτείνουσαν ὑπὸ μίαν τῶν ἴσων γωνιῶν, καὶ τὰς λοιπὰς πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξει καὶ τὴν λοιπὴν γωνίαν τῇ λοιπῇ γωνίᾳ.

If two triangles have two angles equal to two angles respectively, and one side equal to one side, namely, either that adjoining the equal angles, or that subtending one of the equal angles, they will also have the remaining sides equal to the remaining sides, and the remaining angle to the remaining angle.

ἔστω δύο τρίγωνα τὰ ΑΒΓ, ΔΕΖ τὰς δύο γωνίας τὰς ὑπὸ ΑΒΓ, ΒΓΑ δυσὶ ταῖς ὑπὸ ΔΕΖ, ΕΖΔ ἴσας ἔχοντα ἑκατέραν ἑκατέρᾳ, τὴν μὲν ὑπὸ ΑΒΓ τῇ ὑπὸ ΔΕΖ, τὴν δὲ ὑπὸ ΒΓΑ τῇ ὑπὸ ΕΖΔ· ἐχέτω δὲ καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην, πρότερον τὴν πρὸς ταῖς ἴσαις γωνίαις τὴν ΒΓ τῇ ΕΖ· λέγω, ὅτι καὶ τὰς λοιπὰς πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξει ἑκατέραν ἑκατέρᾳ, τὴν μὲν ΑΒ τῇ ΔΕ τὴν δὲ ΑΓ τῇ ΔΖ, καὶ τὴν λοιπὴν γωνίαν τῇ λοιπῇ γωνίᾳ, τὴν ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ. εἰ γὰρ ἄνισός ἐστιν ἡ ΑΒ τῇ ΔΕ, μία αὐτῶν μείζων ἐστίν.

Let there be two triangles ABC, DEF having two angles ABC, BCA equal to two angles DEF, EFD respectively, namely ABC to DEF, and BCA to EFD; and let them also have one side equal to one side, first, that adjoining the equal angles, namely BC to EF; I say that they will also have the remaining sides equal to the remaining sides respectively, namely AB to DE, and AC to DF, and the remaining angle to the remaining angle, namely BAC to EDF. For, if AB is unequal to DE, one of them is greater.

ἔστω μείζων ἡ ΑΒ, καὶ κείσθω τῇ ΔΕ ἴση ἡ ΒΗ, καὶ ἐπεζεύχθω ἡ ΗΓ. ἐπεὶ οὖν ἴση ἐστὶν ἡ μὲν ΒΗ τῇ ΔΕ, ἡ δὲ ΒΓ τῇ ΕΖ, δύο δὴ αἱ ΒΗ, ΒΓ δυσὶ ταῖς ΔΕ, ΕΖ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ· καὶ γωνία ἡ ὑπὸ ΗΒΓ γωνίᾳ τῇ ὑπὸ ΔΕΖ ἴση ἐστίν· βάσις ἄρα ἡ ΗΓ βάσει τῇ ΔΖ ἴση ἐστίν, καὶ τὸ ΗΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ ἴσον ἐστίν, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται, ὑφʼ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν· ἴση ἄρα ἡ ὑπὸ ΗΓΒ γωνία τῇ ὑπὸ ΔΖΕ. ἀλλὰ ἡ ὑπὸ ΔΖΕ τῇ ὑπὸ ΒΓΑ ὑπόκειται ἴση· καὶ ἡ ὑπὸ ΒΓΗ ἄρα τῇ ὑπὸ ΒΓΑ ἴση ἐστίν, ἡ ἐλάσσων τῇ μείζονι· ὅπερ ἀδύνατον.

Let AB be greater, and let BH be made equal to DE, and let HC be joined. Since therefore BH is equal to DE, and BC to EF, the two BH, BC are equal to the two DE, EF, each to each; and the angle HBC is equal to the angle DEF; therefore the base HC is equal to the base DF, and the triangle HBC is equal to the triangle DEF, and the remaining angles will be equal to the remaining angles, namely those subtended by the equal sides; therefore the angle HCB is equal to the angle DFE. But the angle DFE is by hypothesis equal to the angle BCA; therefore the angle BCH is also equal to the angle BCA, the less to the greater; which is impossible.

οὐκ ἄρα ἄνισός ἐστιν ἡ ΑΒ τῇ ΔΕ. ἴση ἄρα.

Therefore AB is not unequal to DE. Therefore it is equal.

ἔστι δὲ καὶ ἡ ΒΓ τῇ ΕΖ ἴση· δύο δὴ αἱ ΑΒ, ΒΓ δυσὶ ταῖς ΔΕ, ΕΖ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ· καὶ γωνία ἡ ὑπὸ ΑΒΓ γωνίᾳ τῇ ὑπὸ ΔΕΖ ἐστιν ἴση· βάσις ἄρα ἡ ΑΓ βάσει τῇ ΔΖ ἴση ἐστίν, καὶ λοιπὴ γωνία ἡ ὑπὸ ΒΑΓ τῇ λοιπῇ γωνίᾳ τῇ ὑπὸ ΕΔΖ ἴση ἐστίν.

And BC is also equal to EF; therefore the two AB, BC are equal to the two DE, EF, each to each; and the angle ABC is equal to the angle DEF; therefore the base AC is equal to the base DF, and the remaining angle BAC is equal to the remaining angle EDF.

ἀλλὰ δὴ πάλιν ἔστωσαν αἱ ὑπὸ τὰς ἴσας γωνίας πλευραὶ ὑποτείνουσαι ἴσαι, ὡς ἡ ΑΒ τῇ ΔΕ· λέγω πάλιν, ὅτι καὶ αἱ λοιπαὶ πλευραὶ ταῖς λοιπαῖς πλευραῖς ἴσαι ἔσονται, ἡ μὲν ΑΓ τῇ ΔΖ, ἡ δὲ ΒΓ τῇ ΕΖ καὶ ἔτι ἡ λοιπὴ γωνία ἡ ὑπὸ ΒΑΓ τῇ λοιπῇ γωνίᾳ τῇ ὑπὸ ΕΔΖ ἴση ἐστίν.

But indeed, again, let the sides subtending the equal angles be equal, as AB to DE; I say again that the remaining sides will also be equal to the remaining sides, namely AC to DF, and BC to EF; and further, the remaining angle BAC is equal to the remaining angle EDF.

εἰ γὰρ ἄνισός ἐστιν ἡ ΒΓ τῇ ΕΖ, μία αὐτῶν μείζων ἐστίν.

For, if BC is unequal to EF, one of them is greater.

ἔστω μείζων, εἰ δυνατόν, ἡ ΒΓ, καὶ κείσθω τῇ ΕΖ ἴση ἡ ΒΘ, καὶ ἐπεζεύχθω ἡ ΑΘ. καὶ ἐπεὶ ἴση ἐστὶν ἡ μὲν ΒΘ τῇ ΕΖ ἡ δὲ ΑΒ τῇ ΔΕ, δύο δὴ αἱ ΑΒ, ΒΘ δυσὶ ταῖς ΔΕ, ΕΖ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ· καὶ γωνίας ἴσας περιέχουσιν· βάσις ἄρα ἡ ΑΘ βάσει τῇ ΔΖ ἴση ἐστίν, καὶ τὸ ΑΒΘ τρίγωνον τῷ ΔΕΖ τριγώνῳ ἴσον ἐστίν, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται, ὑφʼ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν· ἴση ἄρα ἐστὶν ἡ ὑπὸ ΒΘΑ γωνία τῇ ὑπὸ ΕΖΔ. ἀλλὰ ἡ ὑπὸ ΕΖΔ τῇ ὑπὸ ΒΓΑ ἐστιν ἴση· τριγώνου δὴ τοῦ ΑΘΓ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΒΘΑ ἴση ἐστὶ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΒΓΑ· ὅπερ ἀδύνατον.

Let BC be greater, if possible, and let BΘ be made equal to EF, and let AΘ be joined. And since BΘ is equal to EF, and AB to DE, the two AB, BΘ are equal to the two DE, EF, each to each; and they contain equal angles; therefore the base AΘ is equal to the base DF, and the triangle ABΘ is equal to the triangle DEF, and the remaining angles will be equal to the remaining angles, namely those subtended by the equal sides; therefore the angle BΘA is equal to the angle EFD. But the angle EFD is equal to the angle BCA; therefore, of the triangle AΘC, the exterior angle BΘA is equal to the interior and opposite angle BCA; which is impossible.

οὐκ ἄρα ἄνισός ἐστιν ἡ ΒΓ τῇ ΕΖ·

Therefore BC is not unequal to EF; therefore it is equal.

ἴση ἄρα. ἐστὶ δὲ καὶ ἡ ΑΒ τῇ ΔΕ ἴση.

And AB is also equal to DE.

δύο δὴ αἱ ΑΒ, ΒΓ δύο ταῖς ΔΕ, ΕΖ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ· καὶ γωνίας ἴσας περιέχουσι· βάσις ἄρα ἡ ΑΓ βάσει τῇ ΔΖ ἴση ἐστίν, καὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ ἴσον καὶ λοιπὴ γωνία ἡ ὑπὸ ΒΑΓ τῇ λοιπῇ γωνίᾳ τῇ ὑπὸ ΕΔΖ ἴση.

Therefore the two AB, BC are equal to the two DE, EF, each to each; and they contain equal angles; therefore the base AC is equal to the base DF, and the triangle ABC is equal to the triangle DEF, and the remaining angle BAC is equal to the remaining angle EDF.

ἐὰν ἄρα δύο τρίγωνα τὰς δύο γωνίας δυσὶ γωνίαις ἴσας ἔχῃ ἑκατέραν ἑκατέρᾳ καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην ἤτοι τὴν πρὸς ταῖς ἴσαις γωνίαις, ἢ τὴν ὑποτείνουσαν ὑπὸ μίαν τῶν ἴσων γωνιῶν, καὶ τὰς λοιπὰς πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξει καὶ τὴν λοιπὴν γωνίαν τῇ λοιπῇ γωνίᾳ· ὅπερ ἔδει δεῖξαι.

If therefore two triangles have two angles equal to two angles respectively, and one side equal to one side, namely, either that adjoining the equal angles, or that subtending one of the equal angles, they will also have the remaining sides equal to the remaining sides, and the remaining angle to the remaining angle; which was to be proved.

Notes

1.prop.26τὴν πρὸς ταῖς ἴσαις γωνίαις — Meaning "[the side] adjacent to the equal angles". The noun πλευράν (feminine) is omitted. The preposition πρός with the dative means "near to" or "adjoining". Here it refers to the side lying between the two equal angles.
1.prop.26ὑφʼ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν — The relative pronoun ἃς (accusative plural) refers back to the antecedent γωνίαις (angles, dative plural), but is in the accusative because of the preposition ὑπό (governing the accusative, meaning "under"). The intransitive verb ὑποτείνω (subtend, stretch under) takes the subtended angle in the form of ὑπό + accusative, meaning "the angles which the equal sides subtend."
1.prop.26τριγώνου δὴ τοῦ ΑΘΓ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΒΘΑ ἴση ἐστὶ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΒΓΑ — The genitive construction τριγώνου τοῦ ΑΘΓ functions either as a genitive of reference ("with respect to triangle AΘC") or a partitive genitive indicating the whole to which the exterior angle belongs. The dative phrase τῇ ἐντὸς καὶ ἀπεναντίον is governed by the adjective ἴση (equal), meaning "is equal to the interior and opposite [angle] BCA."

Cite this passage

Euclid, Elements §1.prop.26. Humanitext Reader, https://reader.humanitext.ai/en/text/urn:cts:greekLit:tlg1799.tlg001.humanitext-grc2:1.prop.26

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