Romania — NMO Selection Tests for the Balkan and International Mathematical Olympiads

By the condition in the statement, the points $A$, $B$, $P$, and $Q$ lie on some circle $\omega_1$, and the points $C$, $D$, $P$, and $Q$ lie on some circle $\omega_2$. The line $PQ$ is the radical axis of these two circles, and we proceed to prove that $M$ lies on this radical axis.
Let $MA$ meet $\omega_1$ again at $A'$, and let $MD$ meet $\omega_2$ again at $D'$; if, say, $MA$ is tangent to $\omega_1$, then $A' = A$. Then $\angle(MA', A'B) = \angle(AA', A'B) = \angle(AP, PB) = \angle(CP, PD) = \angle(CD', D'D) = \angle(CD', D'M)$, and
$$\begin{align*} \angle(MA', MB) &= \angle(MA, MB) = \angle(MA, MC) + \angle(MC, MB) \\ &= \angle(MB, MD) + \angle(MC, MB) = \angle(MC, MD) \\ &= \angle(MC, MD'), \end{align*}$$
showing that the triangles $MA'B$ and $MD'C$ are similar and have opposite orientations. Hence $MA'/MD' = MB/MC = MD/MA$, so $MA \cdot MA' = MD \cdot MD'$.

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We still have to decide whether such a point $M$ exists, and deal with the exceptional cases. This can be done in several different ways, e.g., by investigating the orientation-changing similarity transformation mapping $A$ to $D$ and $B$ to $C$. We present a more geometrical proof.
Let the diagonals $AC$ and $BD$ cross at $K$. Choose points $F$, on the diagonal $AC$, and $G$, on the diagonal $BD$, so that the configurations $(A, F, K, C)$ and $(D, K, G, B)$ are similar.
If $AD$ and $BC$ are not parallel, then $F \neq K \neq G$. By similarity, it is sufficient to find the point $M$ such that the triangles $MKF$ and $MGK$ are similar and have opposite orientations. If $KF \neq KG$, one may choose $M$ to be the point where $FG$ crosses the tangent at $K$ to the circle $FGK$.
The exceptional cases are $AD \parallel BC$ and $KF = KG$ (in which case $AC = BD$ by similarity). These cases may be considered as limit cases: in the former case, $M$ tends to $K$ and this can be dealt with along the above lines; in the latter, $M$ becomes an ideal point, so the lines $PQ$ are all parallel. In fact, both cases can be dealt with independently and are relatively easy.