The circle passing through the vertices $A, B$, and $D$ of a rectangular trapezoid $ABCD$ ($\angle A = \angle B = 90^\circ$) intersects the extension of the base $BC$ and the extension of the lateral side $CD$ at points $M$ and $N$ respectively, such that $CM: CB = CN: CD = 1: 2$. Find the ratio of the diagonals $BD$ and $AC$ of the trapezoid.
Prove that $B M=D N$.

## Solution

$B C=2 C M, C D=2 C N, 2 C M^{2}=2 C M \cdot B C=C N \cdot C D=2 C N^{2}$, therefore, $C D=2 C N=2 C M$.

By the condition, $B D$ is the diameter of the given circle. Therefore, $\angle B M D=90^{\circ}$. Since $C M: C D=1: 2$, then $\angle C D M=30^{\circ}, A B^{2}=D M^{2}=3 C M^{2}$ $A C^{2}=A B^{2}+B C^{2}=7 C M^{2}, B D^{2}=M D^{2}+M B^{2}=3 C M^{2}+9 C M^{2}=12 C M^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_a7c46f4bbe36942e4ae1g-06.jpg?height=526&width=526&top_left_y=1775&top_left_x=777)

## Answer

$2 \sqrt{3}: \sqrt{7}$.