10.4. Quadrilateral $A B C D$ is inscribed in a circle. The perpendicular to side $B C$, drawn through its midpoint - point $M$, intersects side $A B$ at point $K$. The circle with diameter $K C$ intersects segment $C D$ at point $P(P \neq C)$. Find the angle between the lines $M P$ and $A D$.
Answer: $90^{\circ}$.

Solution. We will prove that the lines $M P$ and $A D$ are perpendicular. Let $\omega$ be the circle constructed with $K C$ as its diameter, then point $M$ lies on $\omega$, since angle $C M K$ is a right angle. Therefore, $\angle C P M = \angle C K M = \alpha$ (they subtend the arc $C M$ of circle $\omega$). Let $T$ be the point of intersection of lines $A D$ and $M P$. We will assume that point $D$ lies on segment $A T$. The other case is considered similarly. Then $\angle T P D = \alpha$ (angles $T P D$ and $C P M$ are vertical). Therefore, to prove that lines $A D$ and $M P$ are perpendicular, we need to prove that $\angle P D T + \alpha = \frac{\pi}{2}$.

If $\angle P D T = \beta$, then $\angle P D A = \pi - \beta \Rightarrow \angle A B C = \beta$, since quadrilateral $A B C D$ is cyclic. Finally, $B K = C K$, because $M K$ is the perpendicular bisector of $B C$. Therefore, $\angle K C M = \angle K B M = \beta$. From the right triangle $K M C$, we get $\alpha + \beta = \frac{\pi}{2}$. The statement is proved.

![](https://cdn.mathpix.com/cropped/2024_05_06_fd161d7747c25e65cee7g-6.jpg?height=972&width=1003&top_left_y=225&top_left_x=672)

Comment. Only one case of the position of point $D$ is considered - no points are deducted.

It is proved that point $M$ lies on $\omega-1$ point.