The bisectors $A M$ and $B N$ of triangle $A B C$ intersect at point $O$. It is known that $A O=\sqrt{3} M O, N O=(\sqrt{3}-$ 1) $B O$. Find the angles of triangle $A B C$.
$A N: N B=N O: B O$.

## Solution

Let $A B=a$. By the property of the angle bisector of a triangle, $A N: A B=N O: B O$. Therefore, $A N=a(\sqrt{3}-1)$.

Similarly, $B M=\frac{a}{\sqrt{3}}$.

If $C M=c$ and $C N=b$, then $\frac{a}{\sqrt{3}}+c=\frac{b}{\sqrt{3}-1}, c \sqrt{3}=a(\sqrt{3}-1)+b$.

Multiplying the first equation by $\sqrt{3}-1$ and subtracting the second from it, we get $c=\frac{2 a}{\sqrt{3}}$. Thus, $B C=a \sqrt{3}$, $A C=C M \cdot A M /{ }_{M B}=2 a$.

Since $A C^{2}=A B^{2}+B C^{2}$, triangle $A B C$ is a right triangle, $\angle B=90^{\circ}, \angle C=30^{\circ}, \angle A=60^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_3ae27f1f0c68451d31a7g-35.jpg?height=365&width=766&top_left_y=229&top_left_x=652)

Answer

$60^{\circ}, 90^{\circ}, 30^{\circ}$.