## Task 3.

In quadrilateral $A B C D$, $\varangle D A B=110^{\circ}, \varangle A B C=50^{\circ}, \varangle B C D=70^{\circ}$. Let $M$ and $N$ be the midpoints of segments $\overline{A B}$ and $\overline{C D}$, respectively. For point $P$ on segment $\overline{M N}$, it holds that $|A M|:|C N|=|M P|:|N P|$ and $|A P|=|C P|$. Determine the measure of $\varangle A P C$.
## Solution.

Let $K$ be the intersection of lines $A D$ and $B C$. Since $A B C D$ is a cyclic quadrilateral, triangles $A K B$ and $C K D$ are similar. Points $M$ and $N$ are the midpoints of the corresponding segments, so $\varangle A K M = \varangle C K N$. It follows that triangles $A K M$ and $C K N$ are also similar.

![](https://cdn.mathpix.com/cropped/2024_05_30_956ddba8588deef4b337g-13.jpg?height=1293&width=1066&top_left_y=1021&top_left_x=455)

Thus, $|K M| : |K N| = |A M| : |C N|$. According to the problem's condition, we have $|K M| : |K N| = |M P| : |N P|$. We conclude that $K P$ is the angle bisector of $\varangle N K M$, which implies that $K P$ is the angle bisector of $\varangle A K C$.

We will show that quadrilateral $A P C K$ is cyclic. Indeed, let point $A^{\prime} \neq A$ on line $K A$ be such that $\left|A^{\prime} P\right| = |A P| = |C P|$. Triangles $K A^{\prime} P$ and $K C P$ are congruent, and quadrilateral $A^{\prime} P C K$ is a kite. Since triangle $A^{\prime} A P$ is isosceles, it follows that $\varangle P A K = \varangle P A A^{\prime} = \varangle A A^{\prime} P = 180^{\circ} - \varangle P A^{\prime} K = 180^{\circ} - \varangle K C P$. Therefore, $A P C K$ is a cyclic quadrilateral.

Since $\varangle A K C = 180^{\circ} - \varangle B A D - \varangle C B A = 20^{\circ}$, it follows that $\varangle C P A = 160^{\circ}$.