66th Belarusian Mathematical Olympiad

Let the segment $XY$ meet the segments $AB$ and $CD$ at $F$ and $G$, respectively. Since $AX\parallel DC$ and $DC\parallel YB$, the triangles $AXF$ and $BYF$ are similar. Therefore we have $AF/FB=AX/YB$. In the same way we conclude that $DG/GC=DY/XC$.



Suppose that the lines $AD$ and $BC$ are parallel. Then the triangles $BEC$ and $AED$ are similar, so that $EB/AE=EC/DE$. Hence, $$\frac{AF}{FB}=\frac{AX}{YB}=\frac{EC}{DE}=\frac{EB}{AE}=\frac{DY}{XC}=\frac{DG}{GC}.$$ By Thales' theorem, $FG\parallel AD$. Thus, we prove that the lines $AD$, $BC$, and $XY$ are pairwise parallel.

Further, let $AD$ and $CB$ intersect; let $Z$ be their intersection point (without loss of generality, let $Z$ be the intersection point of the rays $AD$ and $CB$). Let the line $ZY$ meet the segment $AB$ at $F^{\prime }$. Show that $A{F}^{\prime }/F^{\prime }B=AF/FB$, which yields that the points $F^{\prime }$ and $F$ coincide, i.e., the lines $AD$, $BC$, $XY$ are concurrent, as required. Let $H$ be the intersection point of the lines $DY$ and $CB$, and $I$ be the intersection point of the lines $XC$ and $BY$. Then since $DY\parallel AB$, we have $A{F}^{\prime }/F^{\prime }B=DY/YH$. Since $DY\parallel XC$ and $DG\parallel YB$, the quadrilateral $BECI$ is a parallelogram, and the triangles $BHY$ and $BCI$ are similar. Therefore $$\frac{A{F}^{\prime }}{F^{\prime }B}=\frac{DY}{YH}=\frac{CI}{YH}=[\Delta BHY\sim \Delta BCI]=\frac{BI}{YB}=\frac{AX}{YB}=\frac{AF}{FB},$$ as required.