and join FC, GB.
Because AF is equal to AG [Constr.
and AB to AC, [Hypothesis.
the two sides FA, AC are equal to the two sides GA, AB, each to each; and they contain the angle FAG common to the two triangles AFC, AGB;
therefore the base FC is equal to the base GB, and the triangle AFC to the triangle AGB, and the remaining angles of the one to the remaining angles of the other, each to each, to which the equal sides are opposite, namely the angle ACF to the angle ABG, and the angle AFC to the angle AGB. [I. 4.
And because the whole AF is equal to the whole AG, of which the parts AB, AC are equal, [Hypothesis.
the remainder BF is equal to the remainder CG. [Axiom 3.
And FC was shewn to be equal to GB;
therefore the two sides BF, FC are equal to the two sides CG, GB, each to each;
and the angle BFC was shewn to be equal to the angle CGB; therefore the triangles BFC, CGB are equal, and their other angles are equal, each to each, to which the equal sides are opposite, namely the angle FBC to the angle GCB, and the angle BCF to the angle CBG. [I. 4.
And since it has been shewn that the whole angle ABG is equal to the whole angle ACF,
and that the parts of these, the angles CBG, BCF are also equal;
therefore the remaining angle ABC is equal to the remaining angle ACB, which are the angles at the base of the triangle ABC. [Axiom 3.
And it has also been shewn that the angle FBC is equal to the angle GCB, which are the angles on the other side of the base.
Wherefore, the angles &c. q.e.d.
Corollary. Hence every equilateral triangle is also equiangular.
