[ Auxiliary circle ] [ Inscribed angle subtended by a diameter ]

Diagonal $B D$ of trapezoid $A B C D$ is equal to $m$, and side $A D$ is equal to $n$. Find the base $C D$, given that the base, diagonal, and side of the trapezoid emanating from vertex $C$ are equal to each other.
Points $D, A$ and $B$ lie on a circle with center at point $C$.

## Solution

Since $C D=C A=C B$, points $D, A$ and $B$ lie on a circle with center at point $C$ and radius equal to $C D$. Let $K$ be the second intersection point of line $C D$ with this circle. Then $D K$ is a diameter, $A B \| D K$. Therefore, $B K=A D=n$, and $\angle D B K=90^{\circ}$. By the Pythagorean theorem,

$$
D K=\sqrt{D M^{2}+B K^{2}}=\sqrt{m^{2}+n^{2}} .
$$

Therefore,

$$
C D=\frac{1}{2} D K=\frac{1}{2} \sqrt{m^{2}+n^{2}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_9466e5d28b3947d5c4a5g-09.jpg?height=438&width=531&top_left_y=1959&top_left_x=772)

## Answer

$\frac{1}{2} \sqrt{m^{2}+n^{2}}$

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