Let $\triangle ABC$ be an acute angled triangle and let $k$ be its circumscribed circle. A point $O$ is given in the interior of the triangle, such that $CE=CF$, where $E$ and $F$ are on $k$ and $E$ lies on $AO$ while $F$ lies on $BO$. Prove that $O$ is on the angle bisector of $\angle ACB$ if and only if $AC=BC$.
My solution:
Part a): If $AC=BC$
Since $AC=BC$ and $CE=CF$ $\Longrightarrow$ $\triangle CEA\cong \triangle CFB$ since $O=BE\cap AF$ $\Longrightarrow$ $\angle FAB=\angle FCB=\angle ECA=\angle EBA$ $\Longrightarrow$ $OA=OB$ $\Longrightarrow$ $CO$ is bisector of $\angle ACB$
Part b): If $CO$ is bisector of $\angle ACB$
Since $CO$ is bisector of $\angle ACB$ $\Longrightarrow$ $\angle OCA=\angle OCB$ and is easy to see that: $\angle CAE=\angle FAC=\angle CBF=\angle CBE$ since $CE=CF$ $\Longrightarrow$ $\triangle OCA\cong \triangle OCB$ since $\angle CAO=\angle CBO$ and $\angle OCA=\angle OCB$...