In a triangle with sides $a$ and $b$ and the angle between them $\alpha$, a semicircle is inscribed, with its diameter lying on the third side. Find the radius of the semicircle.
Connect the vertex of the given angle of the triangle with the center of the semicircle.

## Solution

By connecting the vertex of the given angle with the center of the semicircle, we divide the triangle into two triangles with bases $a$ and $b$ and heights equal to $r$ - the radius of the semicircle. The sum of the areas of the obtained triangles is equal to the area of the given triangle, i.e.,

$$
\frac{1}{2} a r + \frac{1}{2} b r = \frac{1}{2} a b \sin \alpha
$$

From this, we find that

$$
r = \frac{a b \sin \alpha}{a + b}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_36c4956010bd91c4b4eeg-19.jpg?height=198&width=491&top_left_y=387&top_left_x=792)

Answer

$\frac{a b \sin \alpha}{a + b}$.