Question 32
Let $I$ be the incenter of triangle $ABC$, and let the line $l$ be the perpendicular bisector of the segment $AI$. Let the point $P$ lie on the circumcircle of triangle $ABC$, and let the line $AP$ meet $l$ at the point $Q$. The point $R$ lies on $l$ and satisfies $\angle IPR = 90^\circ$. Let the line $IQ$ meet the midline of triangle $ABC$ parallel to the side $BC$ at the point $M$.
$$
\text{Prove that } \angle AMR = 90^\circ.
$$
(Note: in a triangle, the segment joining the midpoints of two sides is called a midline.)
$$
\text{Prove that } \angle AMR = 90^\circ.
$$
(Note: in a triangle, the segment joining the midpoints of two sides is called a midline.)