[ Diameter, main properties $\quad]$ [ Ratios of areas of similar figures ]

On the side $B C$ of triangle $A B C$ as a diameter, a circle is constructed, intersecting the segment $A B$ at point $D$. Find the ratio of the areas of triangles $A B C$ and $B C D$, given that $A C=15, B C=20$ and $\angle A B C=\angle$ $A C D$.
Prove that triangle $ABC$ is a right triangle.

## Solution

If the circle intersects side $AC$ at point $K$, then $\angle KCD = \angle DBK$. Therefore, $\angle DBK = \angle DBC$. This means that point $K$ coincides with point $C$. Therefore, $\angle ACB = 90^\circ$ and $CD$ is the altitude of the right triangle $ACB$ drawn from the vertex of the right angle. Then,

$$
CD = \frac{AC \cdot BC}{AB} = 12
$$

Triangle $ABC$ is similar to triangle $CBD$ with a similarity ratio of $\frac{AB}{BC} = \frac{5}{4}$. Therefore, their areas are in the ratio $\left(\frac{5}{4}\right)^2$.

![](https://cdn.mathpix.com/cropped/2024_05_06_cb0f0d605674d837f9abg-07.jpg?height=417&width=495&top_left_y=0&top_left_x=787)

## Answer

$\frac{25}{16}$.