[asy][asy] size(8cm); defaultpen(fontsize(10pt)); pair A = dir(130), B = dir(210), C = dir(330), M = origin, P = extension((A+C)/2,M,B,C), Q = extension((B+C)/2,M,A,C); dot("$A$", A, dir(130)); dot("$B$", B, dir(210)); dot("$C$", C, dir(330)); dot("$Q$", Q, dir(45)); dot("$P$", P, dir(270)); dot("$M$", M, dir(315)); draw(A--B--C--cycle); draw(circumcircle(A,B,M), dotted); draw(A--M--B, dashed); draw(B--Q, dotted); draw(A--P, dotted); [/asy][/asy] Since $ABPMQ$ is cyclic, we must have $$\angle AQB = \angle APB = \angle AMB = 2\angle C \implies \angle CBQ = \angle CAP = \angle C,$$so triangles $BCQ$ and $ACP$ are isosceles, which implies $\overline{QM} \perp \overline{BC}$ and $\overline{PM} \perp \overline{AC}$. Thus, $M$ is the orthocenter of $\triangle CPQ$ and the conclusion follows.

Let CM intersect PW at O Let <MCB=<MBC =x,<ACM=< CAM=y, <MAB=<ABM=z, 2x+2y+2z=180^ο < QPC =<CAB =y+z due to cyclic ABPQ Finally <COP=180^o - <OCP -<IPC = 180^o - (x+y+z) =180^o - 90^o = 90^o