3. Given a trapezoid $A B C D$ with bases $A B$ and $C D$. Let $E$ be the midpoint of side $A B$, $F$ the midpoint of segment $D E$, and $G$ the intersection of segments $B D$ and $C E$. Express the area of trapezoid $A B C D$ in terms of its height $v$ and the length of segment $F G$, assuming that points $A, F, C$ lie on a straight line.

![](https://cdn.mathpix.com/cropped/2024_04_17_1d8fc6c4c0f260a9dad0g-3.jpg?height=517&width=648&top_left_y=1723&top_left_x=710)
SOLUTION. According to the problem, angles $E F D$ and $A F C$ are right angles, so we have (Fig. 3)

$$
\begin{aligned}
& |\Varangle C D F|=|\Varangle A E F| \quad \text { (alternate interior angles), } \\
& |\Varangle C F D|=|\Varangle A F E| \quad \text { (vertical angles). }
\end{aligned}
$$

Furthermore, point $F$ bisects segment $D E$, so $|D F|=|E F|$ and triangles $C D F$ and $A E F$ are congruent by the SAS (Side-Angle-Side) theorem. Therefore, $|C D|=|A E|$, and together with the equality $|A E|=|E B|$, it follows that $E B$ and $D C$ are two congruent and parallel segments. This means that quadrilateral $E B C D$ is a parallelogram. The intersection $G$ of its diagonals therefore bisects each of them. Points $F$ and $G$ are the midpoints of sides $A C$ and $E C$ of triangle $A E C$, so segment $F G$ is its midline and $|A E|=2|F G|$. Therefore:

$$
|A B|=2|A E|=4 d \quad \text { and } \quad|C D|=|A E|=2 d \text {. }
$$

The area of trapezoid $A B C D$ is $S=\frac{1}{2}(|A B|+|C D|) v=3 d v$.

## PRACTICE PROBLEMS: