The triangle $ABC$ is isosceles with apex at $A{}$ and $M,N,P$ are the midpoints of the sides $BC,CA,AB$ respectively. Let $Q{}$ and $R{}$ be points on the segments $BM$ and $CM$ such that $\angle BAQ =\angle MAR.$ The segment $NP{}$ intersects $AQ,AR$ at $U,V$ respectively. The point $S{}$ is considered on the ray $AQ$ such that $SV$ is the angle bisector of $\angle ASM.$ Similarly, the point $T{}$ lies on the ray $AR$ uch that $TU$ is the angle bisector of $\angle ATM.$ Prove that one of the intersection points of the circles $(NUS)$ and $(PVT)$ lies on the line $AM.$ Proposed by Flavian Georgescu