4. Given a right triangle $ABC$ with a right angle at $C$. On its leg $BC$ of length 26, a circle is constructed with $BC$ as its diameter. A tangent $AP$ is drawn from point $A$ to this circle, different from $AC$. The perpendicular $PH$, dropped from point $P$ to segment $BC$, intersects segment $AB$ at point $Q$. Find the area of triangle $BPQ$, given that $BH: CH = 4: 9$.
Answer: 24.

![](https://cdn.mathpix.com/cropped/2024_05_06_bd69057b1c245517edf6g-06.jpg?height=479&width=540&top_left_y=777&top_left_x=815)

Solution. Let $O$ be the center of $\omega$. Note that

$$
B H=\frac{4}{13} B C=8, \quad C H=18, \quad O H=\frac{1}{2} B C-B H=5, \quad P H=\sqrt{O P^{2}-O H^{2}}=12
$$

Right triangles $B H P$ and $O C A$ are similar because

$$
\angle C B P=\frac{1}{2} \angle C O P=\frac{1}{2}\left(180^{\circ}-\angle C A P\right)=90^{\circ}-\angle C A O=\angle C O A
$$

Then

$$
\frac{A C}{B C}=\frac{1}{2} \cdot \frac{A C}{O C}=\frac{1}{2} \cdot \frac{P H}{B H}=\frac{3}{4}
$$

From the similarity of triangles $B H Q$ and $B C A$, we get $Q H=\frac{3}{4} B H=6$. Therefore,

$$
S_{B P Q}=\frac{1}{2} \cdot P Q \cdot B H=\frac{1}{2} \cdot(P H-Q H) \cdot B H=\frac{1}{2} \cdot 6 \cdot 8=24
$$