Answer: $1: 1$.

![](https://cdn.mathpix.com/cropped/2024_05_06_c43b1f7fb831d10d42ecg-05.jpg?height=497&width=622&top_left_y=431&top_left_x=754)

Solution. Since quadrilateral $A B C D$ is cyclic, the equalities $\angle B A C=\angle B D C$ and $\angle B C A=\angle B D A$ hold. Also note that $\angle B C A=\angle B D C$ by the condition and $\angle B A O=\angle B L O$ as angles subtended by the same arc. Then

$$
\angle B C A=\angle B D C=\angle B A C=\angle B A O=\angle B L O .
$$

Therefore, triangle $C L O$ is isosceles, that is, $O C=O L$. Moreover,

$$
\angle A D O=\angle B C A=\angle B D C=\angle O D C \quad \text { and } \quad \angle A K O=\angle A B O=\angle A B D=\angle A C D .
$$

Thus, triangles $D K O$ and $D C O$ are congruent by two angles and a side, from which $O K=O C$. Therefore, $O K=O L$.
3. The diagonals of the inscribed quadrilateral $A B C D$ intersect at point $O$. Let $K$ and $L$ be the points of intersection of the circumcircle of triangle $A O B$ with the lines $A D$ and $B C$ respectively. Find the ratio $O K: O L$, given that $\angle B C A = \angle B D C$.