In triangle $ABC$, the angle at vertex $B$ is $120^{\circ}$. The angle bisector of this angle intersects side $AC$ at point $P$. The external angle bisector from vertex $C$ intersects the line of side $AB$ at point $Q$, and segment $PQ$ intersects side $BC$ at point $R$. What is the measure of angle $PRA$?
From the condition, it follows that the angles marked at point $B$ are equal, their size being $60^{\circ}$ (Figure 1). The line $B Q$ thus bisects the exterior angle of triangle $B C P$ at $B$. The point $Q$, being the intersection of two exterior angle bisectors, is the center of the excircle of triangle $B C P$ that is tangent to the side $B C$ from the outside, so the line $P Q$ bisects the angle $B P C$.

The same line $P Q$ is also an exterior angle bisector in triangle $A B P$. Due to the $60^{\circ}$ angles formed at $B$, the line $B C$ also bisects the exterior angle $P B Q$ of triangle $A B P$. Therefore, the point $R$ is the center of the excircle of triangle $A B P$ that is tangent to the side $B P$ from the outside.

Now consider triangle $A B P$ together with the incenter $I$ of its incircle (Figure 2). The sought angle $P R A$ is clearly equal to the angle $P R I$. Due to the perpendicularity of the interior and exterior angle bisectors at $B$ and $P$, the quadrilateral $I B R P$ is cyclic, and thus, as angles subtending the arc $I P$, $P R I \varangle = B P I \varangle$. Since $B I$ is the interior angle bisector, this latter angle is equal to half of the $60^{\circ}$ angle $A B P \varangle$ given in the problem.

Therefore, the size of the sought angle $P R A$ is $30^{\circ}$.

Based on the solutions of Csilla Dömötör (Győr, Révai M. Gymnasium, 11th grade) and Anna Lovrics (Budapest, Berzsenyi D. Gymnasium, 12th grade)
![](https://cdn.mathpix.com/cropped/2024_05_02_5cd71a14081e37d72109g-1.jpg?height=800&width=682&top_left_y=721&top_left_x=710)