Question 187
In the triangle $ABC$, let us denote $M, N, P$ the midpoints of the sides $BC, CA, AB$ respectively and let $G$ be the centroid of $ABC$. Let the circumcircle of $BGP$ intersect the line $MP$ at a point $K$ different from $P$, and let the circumcircle of $CGN$ intersect the line $MN$ at a point $L$ different from $N$. Prove $|\angle BAK| = |\angle CAL|$.