Equal angle bisectors

Lemma 1:

If two chords of a circle subtend different acute angles at points on the circle, the smaller angle belongs to the shorter chord.

Proof: Two equal chords subtend equal angles at the centre and equal angles (half as big) at suitable points on the circumference. Of two unequal chords, the shorter one would subtend a smaller angle at the centre since it is further away from it than the larger chord.. Consequently, we get a smaller acute angle at the circumference of the circle.

Lemma 2:

If a triangle has two different angles, the smaller angle has the longer bisector.

Proof: Let $ABC$ be a triangle with $b < c$ . Let $BE$ and $CF$ be the angle bisectors at $B$ and $c$ respectively. We wish to prove that $BE > CF$ . Take $E'$ on $BE$ such that $\angle E'CF = \frac {\angle ABC}{2}$ . Since this is equal to $\angle E'BN$ , we have that $F$ , $B$ , $C$ and $E'$ are cyclic.

Since $\angle ABC < \frac {\angle ABC + \angle ACB}{2} < \frac{\angle ABC + \angle ACB + \angle BAC}{2}$ , we get $\angle CBF < \angle E'CB < 90^{\circ}$ .

By Lemma 1, we get $CF < E'B$ . Hence $BE > BE' > CN$ .

Thus we have proven Lemma 2. By taking the contrapositive of this lemma, we get that if two angle bisectors are equal, the triangle has 2 equal angles, so it is isosceles.