Question 120
Points $K$ and $M$ are chosen on the sides $AB$ and $BC$ of triangle $ABC$ in such a way, that $AK = KM = MC$. Let $N$ be the point of intersection of $AM$ and $CK$, $P$ – the feet of perpendicular from point $N$ to the line $KM$, and $Q$ is such point inside segment $KM$, that $MQ = KP$. Prove, that incircle of $KMB$ touches $KM$ at point $Q$.