Since $AD = CD$, $AB = AC$ and $\angle ADC = \angle BAC$, triangles $ADC$ and $BAC$ are similar by case SAS. Segments $CM$ and $CN$ are corresponding medians, so $\frac{CM}{CN} = \frac{CA}{CB}$ and $\angle BCN = \angle ACM \iff \angle BCN + \angle NCA = \angle ACM + \angle NCA \iff \angle BCA = \angle NCM$. Thus, again by case SAS, triangles $CMN$ and $CAB$ are similar, and therefore $CMN$ is an isosceles triangle with $CM = MN$.