On the side $AB$ of the parallelogram $ABCD$ we take the points $X$ and $Y$ such that the points $A$, $X$, $Y$ and $B$ appear in this order. The lines $DX$ and $CY$ intersect at the point $Z$. Suppose that the area of the triangle $\triangle XYZ$ is equal to the sum of the areas of the triangles $\triangle AXD$ and $\triangle CYB$. Prove that the area of the quadrilateral $XYCD$ is equal to $3$ times the area of the triangle $\triangle XYZ$.