7. Let $A C E$ be a triangle with a point $B$ on segment $A C$ and a point $D$ on segment $C E$ such that $B D$ is parallel to $A E$. A point $Y$ is chosen on segment $A E$, and segment $C Y$ is drawn. Let $X$ be the intersection of $C Y$ and $B D$. If $C X=5, X Y=3$, what is the ratio of the area of trapezoid $A B D E$ to the area of triangle $B C D$ ?
Solution:
$$
39 / 25
$$

Draw the altitude from $C$ to $A E$, intersecting line $B D$ at $K$ and line $A E$ at $L$. Then $C K$ is the altitude of triangle $B C D$, so triangles $C K X$ and $C L Y$ are similar. Since $C Y / C X=8 / 5, C L / C K=8 / 5$. Also triangles $C K B$ and $C L A$ are similar, so that $C A / C B=8 / 5$, and triangles $B C D$ and $A C E$ are similar, so that $A E / B D=8 / 5$. The area of $A C E$ is $(1 / 2)(A E)(C L)$, and the area of $B C D$ is $(1 / 2)(B D)(C K)$, so the
ratio of the area of $A C E$ to the area of $B C D$ is $64 / 25$. Therefore, the ratio of the area of $A B D E$ to the area of $B C D$ is $39 / 25$.