Problems

The vertex $A$ of the acute-angled triangle $ABC$ is connected by a segment with the center $O$ of the circumscribed circle. The height $AH$ is drawn from the vertex $A$. Prove that $\mathrm{\angle }BAH=\mathrm{\angle }OAC$.

The vertex $A$ of the acute-angled triangle $ABC$ is connected by a segment with the center $O$ of the circumscribed circle. The height $AH$ is drawn from the vertex $A$. Prove that $\mathrm{\angle }BAH=\mathrm{\angle }OAC$.

From an arbitrary point $M$ lying within a given angle with vertex $A$, the perpendiculars $MP$ and $MQ$ are dropped to the sides of the angle. From point $A$, the perpendicular $AK$ is dropped to the segment $PQ$. Prove that $\mathrm{\angle }PAK=\mathrm{\angle }MAQ$.

On the circle, the points $A,B,C$ and $D$ are given. The lines $AB$ and $CD$ intersect at the point $M$. Prove that $AC\times AD/AM=BC\times BD/BM$.