[ Law of Sines
[Isosceles, Inscribed, and Circumscribed Trapezoids]

In an isosceles triangle $ABC$, a midline $MN$ is drawn parallel to the base $AC$. The radius of the circle circumscribed around the trapezoid $ACMN$ is $\frac{\sqrt{5}}{2}$ times the radius of the circle circumscribed around triangle $ABC$. Find the angles of triangle $ABC$.
Let $A C=a, \angle A B C=\alpha, R-$ be the radius of the circumcircle of triangle $A B C$. By the Law of Sines,

$$
R=\frac{a}{2 \sin \alpha}=\frac{a}{4 \sin \frac{\alpha}{2} \cos \frac{\pi}{2}}
$$

Let $H$ be the projection of the midpoint $M$ of the lateral side $B C$ onto the base $A C$. Then

$$
H C=\frac{1}{2}(A C-M N)=\frac{1}{2}\left(a-\frac{a}{2}\right)=\frac{\frac{a}{4}}{4}, A H=a-\frac{a}{4}=\frac{3}{4} a
$$

From the right triangles СНМ and АНМ, we find that

$$
\begin{aligned}
& M H=C H \tan \angle A C B=\frac{\frac{a}{4}}{4} \cdot \tan\left(\frac{\pi}{2}-\frac{\alpha}{2}\right)=\frac{a}{4} \cot \alpha, \\
& A M=\sqrt{A H^{2}+M H^{2}}=\sqrt{\frac{9}{16} a^{2}+\frac{a^{2}}{16} \cot^{2} \frac{\alpha}{2}}=\frac{a}{4} \sqrt{9+\cot^{2} \frac{\alpha}{2}}
\end{aligned}
$$

Let $R 1$ be the radius of the circumcircle of the isosceles trapezoid $A C M N$. By the Law of Sines,

![](https://cdn.mathpix.com/cropped/2024_05_06_786127613a7b2f99b094g-20.jpg?height=129&width=742&top_left_y=1336&top_left_x=658)

Since $R 1=\frac{\sqrt{5}}{2} R$, we have

$$
\frac{a \sqrt{9+\cot^{2} \frac{\alpha}{2}}}{\sin \frac{\theta}{2}}=\frac{\sqrt{5}}{2} \cdot \frac{a}{4 \sin \frac{\alpha}{2} \cos \frac{\pi}{2}}
$$

From which we find that $\sin^2 \frac{\alpha}{2}=\frac{1}{2}$. Then

$$
\cos \alpha=1-2 \sin^2 \frac{\alpha}{2}=1-2 \cdot \frac{1}{2}=0
$$

Therefore,

$$
\angle A B C=\alpha=\frac{\pi}{2}, \angle B A C=\angle B C A=\frac{\pi}{4} .
$$

## Answer

$\frac{\pi}{2}, \frac{\pi}{4}$.