A circle of radius $R$, passing through the vertices $A, B$ and $C$ of a rectangular trapezoid $A B C D$ ( $\angle A=\angle B=90^{\circ}$ ) intersects the segments $A D$ and $C D$ at points $M$ and $N$, respectively, such that $A M: A D=C N: C D=1: 3$. Find the area of the trapezoid.
## Solution

Since $D C \cdot D N=D A \cdot D M$, then $\frac{2}{3} D C^{2}=\frac{2}{3} D A^{2}$. Therefore, $D C=D A$.

Let $A M=x$. Then $M D=2 x, C D=A D=3 x$. Since point $M$ lies on the circle with diameter $A C$, then $C M \perp A D$. By the Pythagorean theorem,

$$
C M^{2}=A C^{2}-A M^{2}=C D^{2}-D M^{2}
$$

Therefore,

$$
4 R^{2}-x^{2}=9 x^{2}-4 x^{2}
$$

From this, we find that $x=\frac{R \sqrt{2}}{\sqrt{3}}$. Then

$$
C M^{2}=4 R^{2}-\frac{2}{3} R^{2}=\frac{10}{3} R^{2}
$$

Thus,

$$
\begin{gathered}
S_{\mathrm{ABCD}}=\frac{1}{2}(A D+B C) C M=2 x \cdot C M= \\
\quad=\frac{2 R \sqrt{2} \cdot R \sqrt{10}}{\sqrt{3} \cdot \sqrt{3}}=\frac{4 R^{2} \sqrt{5}}{3}
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_df2bbd76ac4dd2f0f20fg-17.jpg?height=477&width=694&top_left_y=1185&top_left_x=682)

## Answer

$\frac{4 \sqrt{5} R^{2}}{3}$