Triangle $A B C$ has no obtuse angles. On the side $A C$ of this triangle, a point $D$ is taken such that $A D=3 / 4 A C$. Find the angle $A$, given that the line $B D$ divides triangle $A B C$ into two similar triangles.
Prove that $B D \perp A C$.

## Solution

If angle $A D B$ is obtuse, then in triangle $C B D$ one of the angles is obtuse. Then in triangle $A B C$ one of the angles is obtuse, which contradicts the condition. Similarly, angle $A D B$ cannot be acute. Therefore, $\angle A D B=90^{\circ}$. If $\angle B C D=\angle B A D$, then triangle $A B C$ is isosceles. Then its height $B D$ is a median, which contradicts the condition. Therefore,

$\angle B C D=\angle A B D$ and $\angle D B C=\angle B A D$, that is, triangle $A B D$ is similar to triangle $B C D$. Therefore, $\angle B=$ $\angle A B D+\angle D B C=\angle A B D+\angle B A D=90^{\circ}$.

$A D: C D=A B^{2}: B C^{2}$, so $A B: B C=\sqrt{3}: 1$, that is, $\angle A=30^{\circ}$.

## Answer

$30^{\circ}$.