4. Given is a rectangle $A B C D$. Point $E$ is the midpoint of segment $\overline{B D}$, and point $F$ is the midpoint of segment $\overline{E D}$. What is the ratio of the area of triangle $\triangle E C F$ to the area of quadrilateral $A B C D$?
First method:

![](https://cdn.mathpix.com/cropped/2024_05_30_494875a1129ddb904645g-09.jpg?height=268&width=514&top_left_y=2379&top_left_x=314)

Let $P$ be the area of triangle $\triangle E C F$.

Point $F$ is the midpoint of segment $\overline{E D}$, so $|D F|=|F E|$, which means the area of triangle $\triangle D F C$ is equal to the area of triangle $\triangle E C F$ because they have bases of equal length and a common height from vertex $C$. Therefore, the area of triangle $\triangle D E C$ is equal to $2 P$.

1 POINT

Point $E$ is the midpoint of segment $\overline{B D}$, so $|D E|=|E B|$, which means the area of triangle $\triangle E B C$ is equal to the area of triangle $\triangle D E C$ because they have bases of equal length and a common height from vertex $C$. Therefore, the area of triangle $\triangle E B C$ is equal to $2 P$, and the area of triangle $\triangle C D B$ is equal to $4 P$. 1 POINT Since $A B C D$ is a rectangle, $|A B|=|C D|$, $|\measuredangle B A D|=|\measuredangle D C B|$, and $|A D|=|C B|$, so by the S-A-S congruence theorem, $\triangle A B D \cong \triangle C D B$.

Therefore, the area of triangle $\triangle A B D$ is equal to $4 P$, and the area of quadrilateral $A B C D$ is equal to $8 P$.

The ratio of the area of triangle $\triangle E C F$ to the area of quadrilateral $A B C D$ is $\frac{P}{8 P}=\frac{1}{8}=0.125$. 1 POINT TOTAL 6 POINTS

Note: Solutions without an explanation of the congruence of triangles should be scored with a maximum of 5 points. If other necessary explanations are missing, score with a maximum of 3 points.