[Similar auxiliary triangles] Trapezoids (other). $\quad]$

The height of trapezoid $A B C D$ is 5, and the bases $B C$ and $A D$ are 3 and 5, respectively. Point $E$ is on side $B C$, with $B E=2$, $F$ is the midpoint of side $C D$, and $M$ is the intersection point of segments $A E$ and $B F$. Find the area of quadrilateral $A M F D$.
Let the line $B F$ intersect the extension of the base $A D$ at point $N$. Then $S_{A M F D}=S_{A M N}-S_{F D N}$.

## Solution

Let the line $B F$ intersect the extension of the base $A D$ at point $N$. From the equality of triangles $B C F$ and $N D F$, it follows that $D N=B C=3$. Triangles $A M N$ and $E M B$ are similar with a ratio of $A N: B E=4$, so the height $M P$ of triangle $A M N$ is 4 times the height $M Q$ of triangle $B M E$. Since $M Q + M E = P Q = 5$, then $M P = 4 / 5 P Q = 4$. Since $F N = B F$, the height $F G$ of triangle $F D N$ is half the height $B H$ of triangle $A B N$, that is,

$F G = 5 / 2$. Therefore, $S_{A M F D} = S_{A M N} - S_{F D N} = 1 / 2 A N \cdot M P - 1 / 2 D N \cdot F G = 1 / 2 \cdot 8 \cdot 4 - 1 / 2 \cdot 3 \cdot 5 / 2 = 49 / 4$.

![](https://cdn.mathpix.com/cropped/2024_05_06_4c124ad32a76c9faf91dg-10.jpg?height=440&width=644&top_left_y=2027&top_left_x=707)

## Answer

$12 \frac{1}{4}$.