The angle at the vertex of an isosceles triangle is $40^{\circ}$. One of the lateral sides serves as the diameter of a semicircle, which is divided into three parts by the other sides. Find these parts.

#
The point of intersection of the circle with the base of the triangle is the base of the height drawn from its vertex.

## Solution

Let $AB$ be the diameter of the circle, $M$ be the point of its intersection with the base $BC$, and $K$ be the point of intersection with the side $AC$. Then $AM$ is the height, and therefore, the bisector of the isosceles triangle $BAC$. Consequently,

$$
\cup BM = \cup MK = 40^{\circ}, \cup KA = 180^{\circ} - 2 \cdot 40^{\circ} = 100^{\circ}
$$

## Answer