[Central Angle. Arc Length and Circumference $]$

In a circle, a trapezoid $A B C D$ is inscribed ($A D$ is the larger base). A perpendicular is drawn from vertex $C$ to $A D$, intersecting the circle at point $E$. The ratio of the length of arc $B C$ (not containing point $D$) to the length of arc $C D E$ is $1: 2$. The radius of the circle is equal to the height of the trapezoid. Find the ratio $A D: B C$.
The sum of the degree measures of the two specified arcs is $180^{\circ}$.

## Solution

Since $\angle B C E=90^{\circ}$, $B E$ is the diameter of the circle. Therefore, the sum of the degree measures of the specified arcs is $180^{\circ}$,

$$
\cup B C=60^{\circ}, \cup C D E=120^{\circ}, \angle B E C=30^{\circ}, \angle C B E=60^{\circ}
$$

Let $R$ be the radius of the circle, $O$ its center, and $B K=C F=R$ the heights of the trapezoid. Then

$$
B C=B E \sin 30^{\circ}=R, C E=2 R \sin 60^{\circ}=R \sqrt{3}
$$

If $M$ and $N$ are the midpoints of the chords $A D$ and $C E$, then

$$
O M=N F=C F-C N=R-\frac{R \sqrt{3}}{2}=\frac{R(2-\sqrt{3})}{2}
$$

Then

$$
\begin{gathered}
A D=2 A M=2 \sqrt{O A^{2}-O M^{2}}= \\
=2 \sqrt{R^{2}-R^{2} \frac{(2-\sqrt{3})^{2}}{4}}=R \sqrt{4 \sqrt{3}-3}
\end{gathered}
$$

Therefore,

$$
\frac{A D}{B C}=\sqrt{4 \sqrt{3}-3}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_df2bbd76ac4dd2f0f20fg-18.jpg?height=463&width=508&top_left_y=1023&top_left_x=775)

Answer

$\sqrt{4 \sqrt{3}-3}$