Lopovok L.M.

The circle constructed on the height $A D$ of the right triangle $A B C$ as a diameter intersects the leg $A B$ at point $K$, and the leg $A C$ at point $M$. The segment $K M$ intersects the height $A D$ at point $L$. It is known that the segments $A K$, $A L$, and $A M$ form a geometric progression (i.e., $\frac{A K}{A L}=\frac{A L}{A M}$). Find the acute angles of triangle $A B C$.
Prove that the quadrilateral $A K D M$ is a rectangle.

## Solution

Since $A D$ and $K M$ are diameters of the given circle, the quadrilateral $A K D M$ is a rectangle. Let $P$ be the projection of vertex $A$ onto the diameter $K M$. Then

$$
A L^{2}=A K \cdot A M=K M \cdot A P=A D \cdot A P=2 A L \cdot A P
$$

From this, we find that $A L=2 A P$. Therefore, in the right triangle $A P L$, the angle $A P L$ is $30^{\circ}$, and since $\angle L A P = 2 \angle L A M$, then $\angle L A M = 15^{\circ}$. Thus,

$$
\angle A C B = 90^{\circ} - \angle D A C = 90^{\circ} - \angle L A M = 75^{\circ}, \angle A B C = 90^{\circ} - 75^{\circ} = 15^{\circ}.
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_e3b1ddf947c7dd2094eeg-30.jpg?height=363&width=946&top_left_y=447&top_left_x=562)

## Answer

$15^{\circ}, 75^{\circ}$.