The angles at the base of the isosceles triangle $A O L$ are $15^{\circ}$.

## Solution

Let $A C < A B$. Since $A L$ is the bisector of angle $C A B$, then $\cup C L = \cup B L$. Therefore, $C L = B L$, and since $O C = O B$, the points $L$ and $O$ lie on the perpendicular bisector of segment $B C$. Therefore, $O L \| A H$.

Let $P$ be the midpoint of $B C$. Then

$$
\begin{gathered}
\angle P O E = 90^{\circ} - \angle P E O = 90^{\circ} - \angle A E H = 90^{\circ} - 60^{\circ} = 30^{\circ}, \\
\angle H A F = \angle P L A = \angle L A O = 15^{\circ}.
\end{gathered}
$$

Since $\angle H A E = 30^{\circ}$, point $F$ lies between points $H$ and $E$.

$$
S_{\triangle \mathrm{OAL}} = \frac{1}{2} A L \cdot \frac{1}{2} A L \cdot \operatorname{tg} 15^{\circ} = \frac{1}{4} A L^{2} \operatorname{tg} 15^{\circ} = 8 \operatorname{tg} 15^{\circ}
$$

$$
S_{\Delta \mathrm{FAE}} = \frac{1}{2} A F \cdot A E \sin 15^{\circ} = \frac{1}{2} \cdot \frac{A H}{\cos 15^{\circ}} \cdot \frac{A H}{\cos 30^{\circ}} \cdot \sin 15^{\circ} = \frac{1}{2} \cdot \frac{2 A H^{2}}{\sqrt{3}} \cdot \operatorname{tg} 15^{\circ} = 2 \operatorname{tg} 15^{\circ}
$$

$$
S_{\mathrm{OEFL}} = S_{\triangle \mathrm{OAL}} - S_{\triangle \mathrm{FAE}} = 6 \operatorname{tg} 15^{\circ}
$$

Therefore,

$$
\frac{S_{\triangle O A L}}{S_{O E F L}} = \frac{8}{6} = \frac{4}{3}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_e1837d72990d2467a29dg-08.jpg?height=486&width=509&top_left_y=1595&top_left_x=777)

## Answer

$\frac{4}{3}$
A circle with center $O$ is inscribed with a triangle $ABC (A>90^{\circ})$. The extension of the angle bisector $AF$ of angle $A$ of this triangle intersects the circle at point $L$, and the radius $AO$ intersects the side $BC$ at point $E$. Let $AH$ be the altitude of triangle $ABC$. Find the ratio of the area of triangle $OAL$ to the area of quadrilateral $OEF L$, given that $AL=4\sqrt{2}$, $AH=\sqrt{2\sqrt{3}}$, and $\angle AEH=60^{\circ}$.