A perpendicular dropped from the vertex of a rectangle to the diagonal divides the right angle in the ratio $1: 3$.

Find the angle between this perpendicular and the other diagonal.
If $M$ is the intersection point of the diagonals $A C$ and $B D$ of rectangle $A B C D$, then triangle $A M D$ is isosceles.

## Solution

Let $A K$ be the perpendicular dropped from vertex $A$ of rectangle $A B C D$ to diagonal $D B$, and $\angle B A K = 3 \angle D A K ; M$ is the intersection point of the diagonals. Then $\angle D A K = 90^{\circ} : 4 = 22.5^{\circ}, \angle A D M = 67.5^{\circ}$.

Since triangle $A M D$ is isosceles, $\angle D A M = \angle A D M = 67.5^{\circ}$. Therefore, $\angle K A M = \angle D A M - \angle D A K = 67.5^{\circ} - 22.5^{\circ} = 45^{\circ}$.

## Answer

$45^{\circ}$.