China Hong Kong Mathematical Olympiad

Let $B'$ be the point such that $AB'BC$ is an isosceles trapezoid. Thus, $B'$ lies on the circle passing through $A$, $B$, $C$, $D$. Also, we have $AP = CB = AB'$ and $CQ = AB = CB'$. Let $E$ and $F$ be the midpoints of $B'P$ and $B'Q$ respectively. Then $AE \perp B'P$ and $CF \perp B'Q$.

First of all, as $\angle PAB' + \angle QCB' = (180^\circ - \angle B'AD) + (180^\circ - \angle B'CD) = 180^\circ$, we know that $\angle EAB' + \angle FCB' = 90^\circ$. This implies the right-angled triangles $AEB'$ and $B'FC$ are similar. Hence, $\frac{AE}{AB'} = \frac{B'F}{B'C}$.

By the midpoint theorem, $B'EMF$ is a parallelogram. Therefore, we have
$$\frac{AE}{EM} = \frac{AE}{B'F} = \frac{AB'}{B'C}.$$
Together with
$$\begin{aligned} \angle AEM &= 90^\circ + \angle B'EM = 270^\circ - \angle EB'F \\ &= 270^\circ - (360^\circ - \angle AB'C - \angle AB'E - \angle CB'F) = \angle AB'C, \end{aligned}$$
we find that $\triangle AEM \sim \triangle AB'C$. This yields $\triangle AEB' \sim \triangle AMC$ by spiral similarity. Therefore, $\angle AMC = \angle AEB' = 90^\circ$.