Macedonian Mathematical Olympiad

Then $\angle ABC>\angle BAC$ i.e. $\angle ADC=\angle ABC>\angle BAC=\angle BDC$. This implies

But $$\angle AOE=2\angle ACE\text{ and }\angle BOE=2\angle BCE.$$ Hence $\angle AOE<\angle BOE$ i.e. $\overline{AE}<\overline{BE}$ ($△ABO$ is isosceles). From Ceva's theorem we have: $$\frac{\overline{AM}}{MC}\cdot \frac{\overline{CN}}{NB}\cdot \frac{\overline{BE}}{EA}=1,$$ and it holds that $\overline{CM}=\overline{CN}$ (by condition). We get that $\frac{\overline{AM}}{BN}=\frac{\overline{EA}}{\overline{BE}}<1$ i.e. we have $$\overline{AC}=\overline{AM}+\overline{CM}=\overline{AM}+\overline{CN}<\overline{BN}+\overline{CN}=\overline{BC}$$ which is in contradiction with (1). Analogously we can show that $\overline{AC}<\overline{BC}$ is impossible. This implies $\overline{AC}=\overline{BC}$.