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. $\overline{AE} < \overline{BE}$
($\triangle ABO$ 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 $\overline{AC} = \overline{BC}$.