Question 49
Let $ABC$ be an acute triangle for which $\angle BAC = 30^\circ$. Take a point $X$ inside of the triangle $ABC$ in such a way that $\angle XBC = \angle XCB = 30^\circ$. Also, take points $P$ and $Q$ on the straight lines $BX$ and $CX$, respectively, in such a way that $AP = BP$ and $AQ = CQ$ are satisfied. Let $M$ be the midpoint of the line segment $BC$. Prove that $\angle PMQ = 90^\circ$ must hold.
