to each, namely, ABC to DEF, and BCA to EFD; and let them have also one side equal to one side; and first let those sides be equal which are adjacent to the equal angles in the two triangles, namely, BC to EF: the other sides shall be equal, each to each, namely, AB to DE, and AC to DF, and the third angle BAC equal to the third angle EDF.
For if AB be not equal to DE, one of them must be greater than the other. Let AB be the greater, and make BG equal to DE, [I. 3.
and join GC.
Then because GB is equal to DE, [Construction.
and BC to EF; [Hypothesis.
the two sides GB, BC are equal to the two sides DE, EF, each to each;
and the angle GBC is equal to the angle DEF; [Hypothesis.
therefore the triangle GBC is equal to the triangle DEF, and the other angles to the other angles, each to each, to which the equal sides are opposite; [I. 4.
therefore the angle GCB is equal to the angle DFE.
But the angle DFE is equal to the angle ACB. [Hypothesis.
Therefore the angle GCB is equal to the angle ACB, [Ax. 1.
the less to the greater; which is impossible.
Therefore AB is not unequal to DE,
that is, it is equal to it;
and BC is equal to EF; [Hypothesis.
therefore the two sides AB, BC are equal to the two sides DE, EF, each to each;
and the angle ABC is equal to the angle DEF; [Hypothesis.
therefore the base AC is equal to the base DF, and the third angle BAC to the third angle EDF. [I. 4.
