Brazilian Math Olympiad

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 $\angle BCN=\angle ACM⟺\angle BCN+\angle NCA=\angle ACM+\angle NCA⟺\angle BCA=\angle NCM$. Thus, again by case SAS, triangles $CMN$ and $CAB$ are similar, and therefore $CMN$ is an isosceles triangle with $CM=MN$.