9.4. Petya says to Vasya: «I constructed a non-isosceles triangle $A B C$ and drew the angle bisectors $A M$ and $C N$. It turned out that $O M=O N$, where $O$ is the point of intersection of the bisectors. Can you determine what the angle $B$ is?» Vasya replies: «No, that can't be, for the segments $O M$ and $O N$ to be equal in a non-isosceles triangle.» Who is right?
Answer: Petya is right. Solution: We will prove that for a non-isosceles triangle with the property \( OM = ON \), the angle \( B \) is uniquely determined, and specifically, \( \angle B = 60^\circ \). Moreover, we will show that any non-isosceles triangle with an angle of \( 60^\circ \) at the vertex has this property. Let the projections of point \( O \) onto sides \( AB \) and \( BC \) be \( L \) and \( K \) respectively. Since \( O \) is the incenter, \( OL = OK \). Because the triangle is non-isosceles, point \( N \) does not coincide with point \( L \) and point \( M \) does not coincide with point \( K \). In the right triangles \( NLO \) and \( MKO \), the equalities \( NO = OM \) and \( LO = OK \) hold, so these two triangles are equal.

![](https://cdn.mathpix.com/cropped/2024_05_06_849b96d295f9f61a85abg-1.jpg?height=43&width=916&top_left_y=1760&top_left_x=250)
In the case of a non-isosceles triangle \( ABC \), the order of points \( N, L, B \) on line \( AB \) differs from the order of points \( K, M, B \) on line \( CB \): for example, if point \( L \) lies inside segment \( NB \), then point \( K \) lies outside segment \( MB \). Indeed, otherwise, from the equality of the corresponding angles of the equal triangles \( NLO \) and \( MKO \), it follows that \( \angle ANO = \angle CMO \) and considering the equality of the vertical angles \( \angle AON = \angle COM \), we have (by the second criterion of triangle equality): \( \triangle ANO = \Delta CMO \Rightarrow \angle A / 2 = \angle C / 2 \), which means that the angles at the base of triangle \( ABC \) are equal and it is isosceles. Thus, we obtain a contradiction, and therefore point \( K \) belongs to segment \( MC \), and from the equality \( \triangle NLO = \Delta MKO \) it follows that \( \angle ANC = \angle AMB \). In this case, we have: \( \angle BNO + \angle BMO = 180^\circ = \angle B + \angle NOM \). But \( \angle NOM = \angle AOC = 180^\circ - (\angle A + \angle C) / 2 = 90^\circ + (\angle B / 2) \). Therefore, \( \angle B + (\angle B / 2) + 90^\circ = 180^\circ \) and thus, \( \angle B = 60^\circ \).

![](https://cdn.mathpix.com/cropped/2024_05_06_849b96d295f9f61a85abg-1.jpg?height=445&width=691&top_left_y=1565&top_left_x=1185)