Let $D$ be the midpoint of the base $BC$ of an isosceles triangle $ABC,$ $E$ be the point at the side $AC$ such that $\angle CDE=60^\circ,$ and $M$ be the midpoint of $DE.$ Prove that $\angle AME=\angle BMD.$
Source: Kyiv mathematical festival 2019
Tags: Kyiv mathematical festival, geometry
Let $D$ be the midpoint of the base $BC$ of an isosceles triangle $ABC,$ $E$ be the point at the side $AC$ such that $\angle CDE=60^\circ,$ and $M$ be the midpoint of $DE.$ Prove that $\angle AME=\angle BMD.$
