## Solution.

Since $\varangle B M A=\varangle B N A=90^{\circ}$, quadrilateral $A B M N$ is cyclic.

The center of the circle circumscribed around quadrilateral $A B C D$ is point $P$.

Segments $\overline{P N}$ and $\overline{P M}$ are radii of this circle, so $|P N|=|P M|=|A P|=13$.

![](https://cdn.mathpix.com/cropped/2024_05_30_5bba13a74780d91c8bdbg-28.jpg?height=677&width=877&top_left_y=1706&top_left_x=521)

Since $Q$ is the midpoint of segment $\overline{M N}$, it follows that $|Q N|=|Q M|=5$. 1 point

Triangle $P N M$ is isosceles, so triangles $N P Q$ and $M P Q$ are right triangles. 1 point

By the Pythagorean theorem, we have $|P Q|=\sqrt{|P M|^{2}-|Q M|^{2}}=\sqrt{13^{2}-5^{2}}=12$.
## Task A-4.6.

Let $M$ and $N$ be the feet of the altitudes from vertices $A$ and $B$ of an acute-angled triangle $ABC$. Let $Q$ be the midpoint of segment $\overline{M N}$, and $P$ be the midpoint of side $\overline{A B}$. If $|M N|=10$ and $|A B|=26$, determine the length of $|P Q|$.