Segment $A B$ is the diameter of a circle, and point $C$ lies outside this circle. Segments $A C$ and $B C$ intersect the circle at points $D$ and $M$ respectively. Find the angle $C B D$, if the areas of triangles $D C M$ and $A C B$ are in the ratio $1: 4$.
Triangle DCM is similar to triangle BCA with a coefficient of $\frac{1}{2}$.

## Solution

Triangles $D C M$ and $B C A$ are similar with a coefficient of $\frac{1}{2}$. Therefore, $B C=2 C D$. Consequently,

$$
\sin \angle C B D=\frac{C D}{B C}=\frac{1}{2}
$$

since angle $C B D$ is acute, $\angle C B D=30^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_cb0f0d605674d837f9abg-06.jpg?height=477&width=458&top_left_y=916&top_left_x=800)

## Answer

$30^{\circ}$.