IMO 2006 Shortlisted Problems

Because $AB\parallel CD$, the relation $AK/KB=DL/LC$ readily implies that the lines $AD$, $BC$ and $KL$ have a common point $S$. Consider the second intersection points $X$ and $Y$ of the line $SK$ with the circles $(ABP)$ and $(CDQ)$, respectively. Since $APBX$ is a cyclic quadrilateral and $AB\parallel CD$, one has $$\angle AXB=180^{\circ }-\angle APB=180^{\circ }-\angle BCD=\angle ABC.$$ This shows that $BC$ is tangent to the circle $(ABP)$ at $B$. Likewise, $BC$ is tangent to the circle $(CDQ)$ at $C$. Therefore $SP\cdot SX=S{B}^2$ and $SQ\cdot SY=S{C}^2$. Let $h$ be the homothety with centre $S$ and ratio $SC/SB$. Since $h(B)=C$, the above conclusion about tangency implies that $h$ takes circle $(ABP)$ to circle $(CDQ)$. Also, $h$ takes $AB$ to $CD$, and it easily follows that $h(P)=Y$, $h(X)=Q$, yielding $SP/SY=SB/SC=SX/SQ$. Equalities $SP\cdot SX=S{B}^2$ and $SQ/SX=SC/SB$ imply $SP\cdot SQ=SB\cdot SC$, which is equivalent to $P$, $Q$, $B$ and $C$ being concyclic.

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Alternative solution.

The case where $P=Q$ is trivial. Thus assume that $P$ and $Q$ are two distinct points. As in the first solution, notice that the lines $AD$, $BC$ and $KL$ concur at a point $S$. Let the lines $AP$ and $DQ$ meet at $E$, and let $BP$ and $CQ$ meet at $F$. Then $\angle EPF=\angle BCD$ and $\angle FQE=\angle ABC$ by the condition of the problem. Since the angles $BCD$ and $ABC$ add up to $180^{\circ }$, it follows that $PEQF$ is a cyclic quadrilateral. Applying Menelaus' theorem, first to triangle $ASP$ and line $DQ$ and then to triangle $BSP$ and line $CQ$, we have $$\frac{AD}{DS}\cdot \frac{SQ}{QP}\cdot \frac{PE}{EA}=1\text{and}\frac{BC}{CS}\cdot \frac{SQ}{QP}\cdot \frac{PF}{FB}=1.$$ The first factors in these equations are equal, as $AB\parallel CD$. Thus the last factors are also equal, which implies that $EF$ is parallel to $AB$ and $CD$. Using this and the cyclicity of $PEQF$, we obtain $$\angle BCD=\angle BCF+\angle FCD=\angle BCQ+\angle EFQ=\angle BCQ+\angle EPQ.$$ On the other hand, $$\angle BCD=\angle APB=\angle EPF=\angle EPQ+\angle QPF,$$ and consequently $\angle BCQ=\angle QPF$. The latter angle either coincides with $\angle QPB$ or is supplementary to $\angle QPB$, depending on whether $Q$ lies between $K$ and $P$ or not. In either case it follows that $P$, $Q$, $B$ and $C$ are concyclic.