On a line $r$, points $A$ and $B$ are marked, and on a line $s$, parallel to $r$, points $C$ and $D$ are marked, such that $A B C D$ is a square. Point $E$ is also marked on the segment $C D$.
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a) What is the ratio between the areas of triangles $A B E$ and $B C D$, if $E$ is the midpoint of $C D$?

b) What is the ratio $\frac{D E}{E C}$, so that the area of triangle $B F E$ is double the area of $D F E$, where $F$ is the intersection of segments $A E$ and $B D$?
Solution

a) Let $2 k$ be the area of square $A B C D$, then the area of triangle $A B E$ is equal to $k$ and the area of triangle $B C D$ is also equal to $k$, therefore, the ratio between the areas is 1.

b) If $\frac{A_{B F E}}{A_{D F E}}=2$, then $\frac{B F}{F D}=2$. If $r / / s$, then $\triangle A B F \sim \triangle D E F$, with a ratio of 2, and consequently, $\frac{A B}{D E}=2$, it follows that $E$ is the midpoint of segment $C D$, that is, $\frac{D E}{E C}=1$.

![](https://cdn.mathpix.com/cropped/2024_05_01_a50b5476db779c3986d5g-24.jpg?height=449&width=769&top_left_y=1158&top_left_x=546)