Question 120
Let $ABC$ be an isosceles triangle with apex at $A$, and let $M, N, P$ be the midpoints of the sides $BC, CA, AB$, respectively. Let $Q$ and $R$ be points inside the segments $BM$ and $CM$, respectively, so that $\angle BAQ = \angle MAR$. The segment $NP$ crosses $AQ$ and $AR$ at $U$ and $V$, respectively. The point $S \neq U$ lies on the half-line $AQ$ emanating from $A$, so that $SV$ is the internal bisector of $\angle ASM$. Similarly, $T \neq V$ is a point on the half-line $AR$ emanating from $A$, so that $TU$ is the internal bisector of $\angle ATM$. Prove that one of the points where the circles $NUS$ and $PVT$ meet lies on the line $AM$.
Flavian Georgescu
Flavian Georgescu