Let ABC and DEF be two triangles having the two angles angle ABC and angle BCA equal to the two angles angle DEF and angle EFD respectively, namely the angle ABC to the angle DEF, and the angle BCA to the angle EFD, and let them also have one side equal to one side, first that adjoining the equal angles, namely line BC equal to line EF.

I say that the remaining sides equal the remaining sides respectively, namely AB equals line DE and line AC equals line DF, and the remaining angle equals the remaining angle, namely the angle BAC equals the angle EDF.

If line AB does not equal line DE, then one of them is greater.

Let line AB be greater. Make line BG equal to line DE, and join line GC.

Since line BG equals line DE, and line BC equals line EF, the two sides line GB and line BC equal the two sides line DE and line EF respectively, and the angle GBC equals the angle DEF, therefore the base GC equals the base DF, the triangle GBC equals the triangle DEF, and the remaining angles equal the remaining angles, namely those opposite the equal sides. Therefore the angle GCB equals the angle DFE. But the angle DFE equals the angle ACB by hypothesis. Therefore the angle BCG equals the angle BCA, the less equals the greater, which is impossible.

Therefore line AB is not unequal to line DE, and therefore equals it.

But line BC also equals line EF. Therefore the two sides line AB and line BC equal the two sides line DE and line EF respectively, and the angle ABC equals the angle DEF. Therefore the base AC equals the base DF, and the remaining angle BAC equals the remaining angle EDF.

Next, let sides opposite equal angles be equal, as line AB equals line DE.

I say again that the remaining sides equal the remaining sides, namely line AC equals line DF and line BC equals line EF, and further the remaining angle BAC equals the remaining angle EDF.

If line BC is unequal to line EF, then one of them is greater.

Let line BC be greater, if possible. Make line BH equal to line EF, and join line AH.

Since line BH equals line EF, and line AB equals line DE, the two sides line AB and line BH equal the two sides line DE and line EF respectively, and they contain equal angles, therefore the base AH equals the base DF, the triangle ABH equals the triangle DEF, and the remaining angles equal the remaining angles, namely those opposite the equal sides. Therefore the angle BHA equals the angle EFD.

But the angle EFD equals the angle BCA, therefore, in the triangle AHC, the exterior angle BHA equals the interior and opposite angle BCA, which is impossible.

Therefore line BC is not unequal to line EF, and therefore equals it.

But line AB also equals line DE. Therefore the two sides line AB and line BC equal the two sides line DE and line EF respectively, and they contain equal angles. Therefore the base AC equals the base DF, the triangle ABC equals the triangle DEF, and the remaining angle BAC equals the remaining angle EDF.

Therefore if two triangles have two angles equal to two angles respectively, and one side equal to one side, namely, either the side adjoining the equal angles, or that opposite one of the equal angles, then the remaining sides equal the remaining sides and the remaining angle equals the remaining angle.

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