Problem 15. The bisectors $A M$ and $B H$ of triangle $A B C$ intersect at point $O$. It is known that $A O: M O = \sqrt{3}$ and $H O: B O = \sqrt{3} - 1$. Find the angles of triangle $A B C$.
Solution. Consider triangle $B O A$ as the image of an isosceles triangle $B O A$ with angle $B O A$ equal to $120^{\circ}$.

Construct such a triangle $B O A$. On the extension of segment $B O$ (beyond point $O$), lay off segment $H O = (\sqrt{3}-1) B O$. On the extension of segment $A O$ (beyond point $O$), lay off segment $O M = A O: \sqrt{3}$. The lines $B M$ and $A H$ intersect at point $C$. Thus, the constructed triangle $A B C$ is the image of the triangle whose angles need to be found. Segments $B H$ and $A M$ are the images of its angle bisectors.

Instrumental measurements on the constructed diagram give the following: $A H: H C \approx 1: 1.73 ; B M: M C \approx 1: 2$. But $\sqrt{3} \approx 1.73$. This allows us to hypothesize: In the triangle $A B C$ under study

\[
\begin{gathered}
A B: B C = A H: H C = 1: \sqrt{3} \\
A B: A C = B M: M C = 1: 2, \text{ i.e., } \\
A B: B C: A C = 1: \sqrt{3}: 2
\end{gathered}
\]

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If the equalities (1) are true, then $A B^{2}: B C^{2}: A C^{2} = 1: 3: 4$, i.e., triangle $A B C$ must be a right triangle ( $\angle B$ is a right angle, $\angle C = 30^{\circ}$ ).

From the above, it is clear that the problem has a unique solution. Therefore, it remains only to prove that triangle $A B C$ is a right triangle, $\angle C = 30^{\circ}$, and $\angle A = 60^{\circ}$. This means it satisfies the condition of the problem.

Let $A M = M C = 1$. Then $A B = 0.5 \sqrt{3} ; B M = 0.5 ; B C = 1.5 ; A C = \sqrt{3}$.

By the property of the angle bisector of a triangle, we get:

\[
A O: O M = A B: B M = \sqrt{3}
\]

Further, $A H: H C = A B: B C = 1: \sqrt{3}$ and $A H = \frac{1}{1+\sqrt{3}} A C = 0.5 \cdot (3 - \sqrt{3})$. Therefore, $O H: O B = A H: A B = \sqrt{3}: 1$.

Thus, it is proven that the triangle $A B C$ under study is a right triangle, with angle $C$ equal to $30^{\circ}$.