If two triangles have the two angles equal to two angles respectively, and one side equal to one side, namely, either the side 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.
Let ABC, DEF be two triangles having the two angles ABC, BCA equal to the two angles DEF, 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 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 the angle BAC to the angle EDF. For, if AB is unequal to DE, one of them is greater. Let AB be greater, and let BG be made equal to DE; and let GC be joined.
Then, since BG is equal to DE, and BC to EF, the two sides GB, BC are equal to the two sides DE, EF respectively; and the angle GBC is equal to the angle DEF;
further the remaining angle BAC is equal to the remaining angle EDF. For, if BC is unequal to EF, one of them is greater. Let BC be greater, if possible, and let BH be made equal to EF; let AH be joined.
Then, since BH is equal to EF, and AB to DE, the two sides AB, BH are equal to the two sides DE, EF respectively, and they contain equal angles;
and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend; [I. 4]
equal to the interior and opposite angle BCA:
therefore the two sides AB, BC are equal to the two sides DE, EF respectively, and they contain equal angles;
EDF. [I. 4] Therefore etc.
Q. E. D. 1 2