Let $S$ be the foot of the perpendicular from $B$ to the line $AQ$, then $SQPB$ is a rectangle. Since $\angle ASB=90$ and $AM=MB$, $M$ lies on the perpendicular bisector of $SB$. Thus, $M$ also lies on the perpendicular bisector of $PQ$ (which is the same line), so $MP=MQ$.

Let $M_1$ be the foot of the perpendicular from $M$ to $QP$. $\measuredangle{AQC}=\measuredangle{MM_1P}=\measuredangle{BPQ} = \frac{\pi}{2} \implies AQ \parallel MM_1 \parallel BP \implies $ the quadrilaterals $AMM_1Q \sim MBPM_1$. Since $\overline{AM}=\overline{MB} \implies AMM_1Q \cong MBPM_1 \implies \overline{QM_1}=\overline{M_1P}$. Since $MM_1 \perp QP \implies \overline{MP}=\overline{MQ}$.