Let $ABCD$ be a given trapezoid ($AD \| BC$), $EF$ be its midline, and $CG$ be its height (see figure). We will prove that $EF = AG$.

The first method. Construct parallelogram $ABCH$.

![](https://cdn.mathpix.com/cropped/2024_05_06_d3cd76ad0dc53ae543bdg-24.jpg?height=394&width=1306&top_left_y=40&top_left_x=382)

Triangle $DCH$ is isosceles, so its height $CG$ coincides with the median. Therefore, $GF$ is the midline of triangle $DCH$ and is parallel to $CH$ (and $AB$). Moreover, $EF \| AG$. Thus, $AEGF$ is a parallelogram.

Remark. The figure shows the case $BC < AD$, but our reasoning does not rely on this restriction.

The second method. Let $AD = a$, $BC = b$, then $DG = \frac{a - b}{2}$. Therefore, $AG = AD - DG = a - \frac{a - b}{2} = \frac{a + b}{2} = EF$. Now, by the Pythagorean theorem, $EF^2 = AG^2 = 25^2 - 15^2 = 10 \cdot 40 = 20^2$.

## Answer

20.
$:$ Folkiore

Find the midline of an isosceles trapezoid if its diagonal is 25 and its height is 15.

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