[ The ratio in which the bisector divides the side.] Pythagorean Theorem (direct and inverse). $\quad]$

In a right triangle $ABC$, the bisector $AP$ of the acute angle $A$ is divided by the center $O$ of the inscribed circle in the ratio

$AO: OP=(\sqrt{3}+1):(\sqrt{3}-1)$. Find the acute angles of the triangle.
Let $A P$ be the bisector of triangle $A B C$ with hypotenuse $A B$. By the property of the bisector, $A C: C P=A O: O P=(\sqrt{3}+1):(\sqrt{3}-1)$, so we can assume that $C P=\sqrt{3}-1, A C=\sqrt{3}+1$.

Since $B A: B P=A O: O P=(\sqrt{3}+1):(\sqrt{3}-1)$, let $A B=(\sqrt{3}+1) x, B P=(\sqrt{3}-1) x$.

By the Pythagorean theorem $((\sqrt{3}-1)(x+1))^{2}+(\sqrt{3}+1)^{2}=((\sqrt{3}+1) x)^{2}$, or $(\sqrt{3}-1)^{2}(x+1)=(\sqrt{3}+1)^{2}(x-$

1), that is

$(2-\sqrt{3})(x+1)=(2+\sqrt{3})(x-1)$, from which $x=\frac{2}{\sqrt{3}}, \frac{A C}{A B}=\frac{1}{x}=\frac{\sqrt{3}}{2}$. Therefore, $\angle A=30^{\circ}$.

## Answer

$30^{\circ}, 60^{\circ}$.