In the right-angled triangle $ABC$ at $C$, the angle bisector from vertex $B$ intersects side $AC$ at point $P$ and the circumcircle of the triangle at point $Q$. What are the angles of the triangle if $BP = 2PQ$?
Reflect the triangle $A B C$ over the line $B P$. Use the notations from the diagram. Since $B P$ is the angle bisector, the reflection of $A$, denoted as $A^{\prime}$, lies on $B C$. According to Thales' theorem, $\angle A Q B = 90^{\circ}$. Due to the reflection, $Q$ is the midpoint of the segment $A A^{\prime}$. Therefore, $B Q$ is the median of triangle $A A^{\prime} B$, and from the condition $B P = 2 P Q$, it follows that $P$ is the centroid of this triangle. Consequently, both $B Q$ and $A C$ are symmetry axes of triangle $A A^{\prime} B$, and thus $A B = A^{\prime} B = A^{\prime} A$, making the triangle equilateral. From this, it follows that the angles of triangle $A B C$ are $30^{\circ}, 60^{\circ}$, and $90^{\circ}$.

Kurcz Éva (Szolnok, Tiszaparti Gimn. és Humán SZKI., 9th grade)

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