The bisector of one of the acute angles of a right triangle, divided by the height dropped to the hypotenuse, has a ratio of

$1+\sqrt{2}$, measured from the vertex. Find the acute angles of the triangle.
Consider similar triangles.

## Solution

Let the bisector $A M$ of the right triangle $A B C$ and the height $C H$, drawn from the vertex of the right angle, intersect at point $K$, then $\frac{A K}{K M}=1+\sqrt{2}$. Right triangles $A H K$ and $A C M$ are similar. Therefore, $\cos \angle A=\frac{A H}{A C}=\frac{A K}{A M}=\frac{1+\sqrt{2}}{2+\sqrt{2}}=\frac{1}{\sqrt{2}}$. Hence, $\angle A=45^{\circ}$.

## Answer

$45^{\circ}, 45^{\circ}$.