Macedonian Junior Mathematical Olympiad

From $\overline{CE}=\overline{CF}$ it follows that $\angle CAE=\angle CBF$ (1), as inscribed angles subtending equal chords.

Let us assume first that the triangle is isosceles. Then, from the fact that $O$ lies in the interior of $ABC$ and (1), it follows that $\angle BAO=\angle BAC-\angle CAO=\angle ABC-\angle CBO=\angle ABO$, from where it follows that the triangle $ABO$ is isosceles with base $\overline{AB}$, i.e. $\overline{AO}=\overline{BO}$ (2). From the fact that the triangle $ABC$ is isosceles, it follows that $\overline{AC}=\overline{BC}$ (3). From (1), (2) and (3) it follows that $\angle AOC\cong \angle BOC$, from where we have $\angle ACO=\angle BCO$, i.e. $O$ lies on the bisector of the angle at the vertex $C$.

Remark: $\angle AOC\cong \angle BOC$ does not follow directly from $\angle CAE=\angle CBF$, $\overline{AC}=\overline{BC}$ and $\overline{CO}$ is a common side.

Let's assume now that point $O$ lies on the bisector of the angle at the vertex $C$ and let $M$ and $N$ be the feet of the perpendiculars from $O$ to the sides $AC$ and $BC$ respectively. The right-angled triangles $CON$ and $COM$ are congruent because $\angle ACO=\angle BCO$ and $\overline{CO}$ is a common side, so $\overline{CN}=\overline{CM}$ (4) and $\overline{ON}=\overline{OM}$ (5). The right-angled triangles $BON$ and $AOM$ are congruent from (1) and (5), so $\overline{BN}=\overline{AM}$ (6). By adding (4) and (6) we get that $\overline{AC}=\overline{BC}$ (the points $M$ and $N$ lie in the interior of the sides, since the triangle is acute).