Find the ratio in which the bisector of the angle at the base divides the height.

## Solution

Let $C M$ be the height of the given triangle $A B C$, $C M=20$, $A C=B C$, and $O$ be the center of the inscribed circle. Then $O M$ is the radius of this circle.

Since $A O$ is the bisector of angle $C A B$, then $\frac{O M}{O C}=\frac{A M}{A C}$, and since $\frac{A M}{A C}=\frac{2}{3}$, then

$$
O M=\frac{2}{5} C M=8
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_02603d12f3437ea52c37g-15.jpg?height=381&width=352&top_left_y=1656&top_left_x=859)

## Answer

8.
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$[$ Inscribed and Circumscribed Circles $]$

In an isosceles triangle, the height is 20, and the base is to the lateral side as $4: 3$. Find the radius of the inscribed circle.

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