SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Let $O$ be the center of $\omega$ and $Q$ the intersection point of $CO$ and $\omega$. Since $QD$ and $BM$ are perpendicular to $CD$ then $BM\parallel QD$. Because $N$ is the midpoint of $BM$, which implies that $D(M,B,N,Q)=-1$. Consider this harmonic quartet with circle $\omega$, one has $CPBQ$ is a harmonic quadrilateral. This implies that $A,P,Q$ are collinear and $AP\perp CP$.



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