49th 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 $PE\ne QE$, 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.

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Alternative solution.

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 $(EAD)$, $X{E}^2=XA\cdot XD$. Also, $XA\cdot XD=XP\cdot XQ$ because $P,Q,D,A$ lie on a circle. Therefore $X{E}^2=XP\cdot XQ$.

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 $XP\ne XQ$. The proof is now complete.