Problem

Source: czechoslovak national mo round

Tags: geometry



In triangle $ABC$ let $N, M, P$ be the midpoints of the sides $BC, CA, AB$ and $G$ be the centroid of this triangle. Let the circle circumscribed to $BGP$ intersect the line $MP$ in point $K$, $P \neq K$, and the circle circumscribed to $CGN$ intersect the line $MN$ in point $L$, $N \neq L$. Prove that $ \angle BAK = \angle CAL $.