A circle constructed on the smaller lateral side of a right trapezoid as its diameter touches the larger lateral side, which is equal to $a$.

Find the midline of the trapezoid.
Let the center $O$ of the circle lie on the shorter lateral side $AB$ of the right trapezoid $ABCD$, and $M$ be the midpoint of the longer lateral side $CD$.

![](https://cdn.mathpix.com/cropped/2024_05_06_198acf192511df7a04ddg-38.jpg?height=437&width=374&top_left_y=633&top_left_x=842)

The center of the inscribed circle in an angle lies on the angle bisector, so $CO$ and $DO$ are the bisectors of angles $BCD$ and $ADC$, the sum of which is $180^{\circ}$. Therefore, $\angle COD = 180^{\circ} - (\angle OCD + \angle ODC) = 90^{\circ}$.

The midline $OM$ of trapezoid $ABCD$ is the median of the right triangle $COD$ drawn from the vertex of the right angle, hence,

$OM = \frac{1}{2} CD$.

## Answer

$\frac{a}{2}$.