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In trapezoid $A B C D(A D \| B C)$, from point $E$ - the midpoint of $C D$, a perpendicular $E F$ to line $A B$ is drawn. Find the area of the trapezoid if $A B=5, E F=4$.
It is sufficient to prove that the area of the trapezoid is twice the area of triangle $ABE$ (which is 10). The first method. Draw a line through point $E$ parallel to the side $AB$, and mark the points $P$ and $Q$ where it intersects the lines $AD$ and $BC$ respectively (see figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_82cd6cdb4fc29ee56532g-81.jpg?height=443&width=600&top_left_y=2088&top_left_x=726)

$ABQP$ is a parallelogram. Moreover, triangles $PED$ and $QEC$ are equal (by side and two adjacent angles), so their areas are equal. Thus, $S_{ABCD}=S_{ABQP}=2S_{ABE}$.

The second method. Let $G$ be the intersection point of the lines $BE$ and $AD$. Triangles $BEC$ and $GED$ are equal, so $S_{ABCD}=S_{ABG}=2S_{ABE}$.

The third method. See the solution to problem 54964.