Question 163
Given quadrilateral $ABCD$, such that $AB = BC$. $K$ is the midpoint of $CD$, rays $BK$ and $AD$ intersect at $M$, the circumcircle of $\triangle ABM$ intersects line $AC$ for the second time at point $P$. Show that $\angle BKP = 90^\circ$.
(Anton Trygub)
(Anton Trygub)