Next, let sides which are opposite to equal angles in each triangle be equal to one another, namely, AB to DE: likewise in this case the other sides shall be equal, each to each, namely, BC to EF, and AC to DF, and also the third angle BAC equal to the third angle EDF.
For if BC be not equal to EF, one of them must be greater than the other.
Let BC be the greater, and make BH equal to EF, [I. 3.
and join AH.
Then because BH is equal to EF, [Construction.
and AB to DE; [Hypothesis.
the two sides AB, BH are equal to the two sides DE, EF, each to each;
and the angle ABH is equal to the angle DEF; [Hypothesis.
therefore the triangle ABH 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 BHA is equal to the angle EFD.
But the angle EFD is equal to the angle BCA. [Hypothesis.
Therefore the angle BHA is equal to the angle BCA; [Ax. 1.
that is, the exterior angle BHA of the triangle AHC is equal to its interior opposite angle BCA;
which is impossible. [I. 16.
Therefore BC is not unequal to EF,
that is, it is equal to it;
and AB is equal to DE; [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.
Wherefore, if two triangles &c. q.e.d.
