On the leg $A C$ of the right triangle $A B C$, a circle is constructed with $A C$ as its diameter, which intersects the hypotenuse $A B$ at point $K$. Find the area of triangle $C K B$, if the leg $A C$ is equal to $b$, and the angle $A B C$ is equal to $\beta$.
Prove that $CK$ is perpendicular to $AB$.

## Solution

Point $K$ lies on the circle with diameter $AC$. Therefore, $CK$ is perpendicular to $AB$. From the right triangles $AKC$ and $BKC$, we find that

$$
CK = AC \cos \angle ACK = b \cos \beta, \quad BK = CK \cdot \cot \angle B = b \cos \beta \cdot \cot \beta
$$

Therefore,

$$
\begin{gathered}
S_{\triangle CKB} = \frac{1}{2} CK \cdot BK = \frac{1}{2} b \cos \beta \cdot b \cos \beta \cdot \cot \beta = \\
= \frac{1}{2} b^2 \cdot \cos^2 \beta \cdot \cot \beta
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_1b17058527caf5b5d058g-09.jpg?height=477&width=506&top_left_y=1248&top_left_x=776)

## Answer

$\frac{1}{2} b^2 \cos^2 \beta \cot \beta$