4. On the segment $A B$ of length 10, a circle $\omega$ is constructed with $A B$ as its diameter. A tangent to $\omega$ is drawn through point $A$, and a point $K$ is chosen on this tangent. A line through point $K$, different from $A K$, is tangent to the circle $\omega$ at point $C$. The height $C H$ of triangle $A B C$ intersects segment $B K$ at point $L$. Find the area of triangle $C K L$, given that $B H: A H=1: 4$.
Answer: 8.

![](https://cdn.mathpix.com/cropped/2024_05_06_bd69057b1c245517edf6g-02.jpg?height=483&width=648&top_left_y=478&top_left_x=767)

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

$$
B H=\frac{1}{5} A B=2, \quad A H=8, \quad O H=\frac{1}{2} A B-B H=3, \quad C H=\sqrt{O C^{2}-O H^{2}}=4
$$

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

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

Then

$$
\frac{A B}{A K}=2 \cdot \frac{A O}{O K}=2 \cdot \frac{B H}{C H}=1
$$

from which $\angle A B K=45^{\circ}$ and $L H=B H=2$. Therefore,

$$
S_{C K L}=\frac{1}{2} \cdot C L \cdot A H=\frac{1}{2} \cdot(C H-L H) \cdot A H=\frac{1}{2} \cdot 2 \cdot 8=8
$$