International Mathematical Olympiad Spain

Let $F$ be the point on the line $AD$ such that $EF \parallel PA$. By hypothesis, the quadrilateral $PQDA$ is cyclic. So if $F$ lies between $A$ and $D$ then $\angle EFD = \angle PAD = 180^{\circ} - \angle EQD$; the points $F$ and $Q$ are on distinct sides of the line $DE$ and we infer that $EFDQ$ is a cyclic quadrilateral. And if $D$ lies between $A$ and $F$ then a similar argument shows that $\angle EFD = \angle EQD$; but now the points $F$ and $Q$ lie on the same side of $DE$, so that $EDFQ$ is a cyclic quadrilateral.

In either case we obtain the equality $\angle EFQ = \angle EDQ = \angle PAE$ which implies that $FQ \parallel AE$. So the triangles $EFQ$ and $PAE$ are either homothetic or parallel-congruent. More specifically, triangle $EFQ$ is the image of $PAE$ under the mapping $f$ which carries the points $P, E$ respectively to $E, Q$ and is either a homothety or translation by a vector. Note that $f$ is uniquely determined by these conditions and the position of the points $P, E, Q$ alone.

Let now $G$ be the point on the line $BC$ such that $EG \parallel PB$. The same reasoning as above applies to points $B, C$ in place of $A, D$, implying that the triangle $EGQ$ is the image of $PBE$ under the same mapping $f$. So $f$ sends the four points $A, P, B, E$ respectively to $F, E, G, Q$.

If $PE \neq QE$, so that $f$ is a homothety with a centre $X$, then the lines $AF, PE, BG$—i.e. the lines $AD, PQ, BC$—are concurrent at $X$. And since $PQDA$ and $QPBC$ are cyclic quadrilaterals, the equalities $XA \cdot XD = XP \cdot XQ = XB \cdot XC$ hold, showing that the quadrilateral $ABCD$ is cyclic.

Finally, if $PE = QE$, so that $f$ is a translation, then $AD \parallel PQ \parallel BC$. Thus $PQDA$ and $QPBC$ are isosceles trapezoids. Then also $ABCD$ is an isosceles trapezoid, hence a cyclic quadrilateral.

Here is another way to reach the conclusion that the lines $AD$, $BC$ and $PQ$ are either concurrent or parallel. From the cyclic quadrilateral $PQDA$ we get
$$\angle PAD = 180^{\circ} - \angle PQD = \angle QDE + \angle QED = \angle PAE + \angle QED.$$
Hence $\angle QED = \angle PAD - \angle PAE = \angle EAD$. This in view of the tangent-chord theorem means that the circumcircle of triangle $EAD$ is tangent to the line $PQ$ at $E$. Analogously, the circumcircle of triangle $EBC$ is tangent to $PQ$ at $E$.

Suppose that the line $AD$ intersects $PQ$ at $X$. Since $XE$ is tangent to the circle $(EAD)$, $XE^2 = XA \cdot XD$. Also, $XA \cdot XD = XP \cdot XQ$ because $P, Q, D, A$ lie on a circle. Therefore $XE^2 = XP \cdot XQ$.

It is not hard to see that this equation determines the position of the point $X$ on the line $PQ$ uniquely. Thus, if $BC$ also cuts $PQ$, say at $Y$, then the analogous equation for $Y$ yields $X = Y$, meaning that the three lines indeed concur. In this case, as well as in the case where $AD \parallel PQ \parallel BC$, the concluding argument is the same as in the first solution.

It remains to eliminate the possibility that e.g. $AD$ meets $PQ$ at $X$ while $BC \parallel PQ$. Indeed, $QPBC$ would then be an isosceles trapezoid and the angle equality $\angle PBE = \angle QCE$ would force that $E$ is the midpoint of $PQ$. So the length of $XE$, which is the geometric mean of the lengths of $XP$ and $XQ$, should also be their arithmetic mean—impossible, as $XP \neq XQ$. The proof is now complete.