Question 84
Let $ABCD$ be a square, $M$ be the midpoint of the side $AD$, $T$ be the common point of the lines $BM$ and $CD$, and $CP \perp BM$, $P \in MB$. The perpendicular line from $A$ to $AP$ intersects the line $BM$ in $Q$. Prove that:
a) $\angle APQ = \angle PCQ = 45^\circ$;
b) $PQ = QT = PC$.
a) $\angle APQ = \angle PCQ = 45^\circ$;
b) $PQ = QT = PC$.