Inside the acute-angled triangle $ABC$, mark the point $O$ so that $\angle AOB=90^o$, a point $M$ on the side $BC$ such that $\angle COM=90^o$, and a point $N$ on the segment $BO$ such that $\angle OMN = 90^o$. Let $P$ be the point of intersection of the lines $AM$ and $CN$, and let $Q$ be a point on the side $AB$ that such $\angle POQ = 90^o$. Prove that the lines $AN, CO$ and $MQ$ intersect at one point.