Fishing for Some Angles

Let $\angle ABM = \angle MBC = a$ and $\angle CMQ = \angle QMP = b$ . Because quadrilateral $BEDC$ is cyclic, we have that $\angle ACP = a$ and $\angle ABM = \angle MPC = \angle ACM = a$ because they substend the same arc or arc with the same degree.

Now notice that $Q$ is the incenter of $\triangle PMC$ because it is the intersection of two angle bisectors. This implies that $\angle MPC = \angle QPC = \frac{a}{2}$ .

$\angle ACP = \angle AMP = a$ since they substend the same arc. However this implies that $\angle AMP = \angle MPC = a \implies AM || PC$ which implies that quadrilateral $AMCP$ is an isosceles trapezoid.

This implies that $MC = AP$ and since $PQ = MC$ , we have $PQ = AP$ which implies $\triangle APQ$ is isosceles. This gives us $\angle AQP = \angle PAQ$ and since $\angle PAC = \angle PMC = 2b$ because they substend the same arc, we have that $\angle AQP = \angle PAQ = 2b$ .

Call the intersection of the diagonals of $AMCP$ , $T$ . Since $\angle MAC = \angle MPC = a$ and $\angle AMP = \angle ACP = a$ , we have that $\angle MTC = 2a$ by exterior angle theorem which further implies that $\angle PTQ = 180^{\circ} - 2a$ . Now, equating the sum of the angles of $\triangle PTQ$ and $\triangle TMC$ to $180^{\circ}$ we get the following two equations: $\begin{aligned} 180^{\circ} - 2a + 2b + \frac{a}{2} & = 180^{\circ} \\ 2b + a + 2a & = 180^{\circ} \end{aligned}$ which can be solved easily to get $a = 40^{\circ}$ which implies that $\angle ABC = 80^{\circ}$ .