8,9

The acute angle of an isosceles trapezoid is $75^{\circ}$. Lines passing through the ends of one of the bases of the trapezoid, parallel to the opposite lateral sides, intersect on the circle circumscribed around the trapezoid. Find the ratio of the bases of the trapezoid.
Prove that the specified lines cannot pass through the ends of the larger base of the trapezoid. Apply the generalized sine theorem $(a=2 R \sin \boldsymbol{\alpha})$.

## Solution

Let the specified lines pass through the ends $A$ and $D$ of the larger base $A D$ of the given trapezoid $A B C D$ and intersect at point $M$, which is located on the circumscribed circle of trapezoid $A B C D$. Since $A M \| C D$, then $\angle M D A = \angle B A D = 75^{\circ}$. Similarly, $\angle M A D = 75^{\circ}$. Then the angle $\angle A M D$ of triangle $A M D$ is

$$
180^{\circ} - 2 \cdot 75^{\circ} = 30^{\circ},
$$

and since points $M$ and $B$ lie on opposite sides of the chord $A D$, then

$$
\angle A B D = 180^{\circ} - \angle A M D = 150^{\circ},
$$

which is impossible, since angle $A B D$ is part of angle $A B C$, which equals $105^{\circ}$.

Now let the specified lines pass through the ends $B$ and $C$ of the smaller base of the given trapezoid and intersect at point $M$, which is located on the circumscribed circle of this trapezoid. Then

$$
\begin{gathered}
\angle B M C = 180^{\circ} - (\angle M B C + \angle M C B) = 180^{\circ} - (\angle A D C + \angle B A D) = \\
= 180^{\circ} - 150^{\circ} = 30^{\circ}.
\end{gathered}
$$

Since points $M$ and $A$ lie on the same side of the chord $B D$, then

$$
\angle B M D = \angle B A D = 75^{\circ}.
$$

Thus,

$$
\begin{gathered}
\angle A M B = \angle C M D = \angle B M D - \angle B M C = 75^{\circ} - 30^{\circ} = 45^{\circ}, \\
\angle A M D = \angle A M B + \angle B M C + \angle C M D = 30^{\circ} + 45^{\circ} + 45^{\circ} = 120^{\circ}.
\end{gathered}
$$

Let $R$ be the radius of the circumscribed circle of the given trapezoid. Then

$$
\begin{gathered}
B C = 2 R \sin \angle B M C = 2 R \sin 30^{\circ} = R \\
A D = 2 R \sin \angle A M D = 2 R \sin 120^{\circ} = R \sqrt{3}
\end{gathered}
$$

Therefore, $\frac{A D}{B C} = \sqrt{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9466e5d28b3947d5c4a5g-14.jpg?height=477&width=472&top_left_y=1442&top_left_x=793)

## Answer

![](https://cdn.mathpix.com/cropped/2024_05_06_9466e5d28b3947d5c4a5g-14.jpg?height=72&width=103&top_left_y=2012&top_left_x=14)