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In an isosceles trapezoid, the height is 10, and the diagonals are perpendicular to each other. Find the midline of the trapezoid.
Through the vertex of the smaller base of the trapezoid, draw a line parallel to the diagonal.

## Solution

First method.

Through the vertex $C$ of the smaller base $BC$ of trapezoid $ABCD$, draw a line parallel to diagonal $BD$. Let $M$ be the point of intersection of this line with the extension of base $AD$, and $CK$ be the height of the trapezoid.

Since $CM = BD = AC$ and $\angle ACM = 90^{\circ}$, then $\angle CMK = 45^{\circ}$ and $KM = CK$. Therefore, the midline of the trapezoid is equal to

$$
\frac{BC + AD}{2} = \frac{AD + DM}{2} = \frac{1}{2} AM = MK = CK = 10
$$

Second method.

Let $P$ and $Q$ be the midpoints of the bases $BC$ and $AD$, respectively, and $R$ and $S$ be the midpoints of the lateral sides $AB$ and $CD$ of trapezoid $ABCD$. The sides of quadrilateral $PSQR$ are parallel to the diagonals of the trapezoid as the midlines of the corresponding triangles, so $PSQR$ is a rectangle. Its diagonal $PQ$ is equal to the height of the trapezoid, and diagonal $RS$ is the midline of the trapezoid. Since the diagonals of a rectangle are equal, then $RS = PQ = 10$.

![](https://cdn.mathpix.com/cropped/2024_05_06_d1936174aeda27dcc10dg-23.jpg?height=383&width=826&top_left_y=966&top_left_x=616)

## Answer

10.

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