[ Relationships between the sides and angles of triangles (other).]

In a right triangle $ABC$ with a right angle at $A$, a circle is constructed on the altitude $AD$ as a diameter, intersecting side $AB$ at point $K$ and side $AC$ at point $M$. Segments $AD$ and $KM$ intersect at point $L$. Find the acute angles of triangle $ABC$ if it is known that $AK: AL = AL: AM$.
Clearly, $A K D M$ is a rectangle and $L$ is the point of intersection of its diagonals. Since $A D \perp B C$ and $A M \perp B A$, we have $\angle D A M = \angle A B C$. Similarly, $\angle K A D = \angle A C B$. Drop a perpendicular $A P$ from point $A$ to line $K M$. Let's assume for definiteness that $\angle B < \angle C$. Then point $P$ lies on segment $K L$. From the similarity of triangles $A K P$ and $M K A$, we get $A K : A P = M K : M A$. Therefore, $A K \cdot A M = A P \cdot M K = A P \cdot A D = 2 A P \cdot A L$. By the condition $A L^2 = A K \cdot A M$, it follows that $A L = 2 A P$, i.e., $\angle A L P = 30^{\circ}$. Clearly, $\angle K M A = \angle A L P / 2 = 15^{\circ}$. Therefore, the acute angles of triangle $A B C$ are 15 and $75^{\circ}$.