Seregay I.n.

Inside triangle $A B C$, a point $O$ is taken such that $\angle A B O=\angle C A O, \angle B A O=\angle B C O, \angle B O C=90^{\circ}$. Find the ratio $A C: O C$.
Let's construct point $D$, symmetric to point $C$ with respect to the line $BO$ (see the figure). Then $\angle BDO = \angle BCO = \angle BAO$. Therefore, point $D$ lies on the circumcircle of triangle $ABO$, and $\angle ADO = \angle ABO = \angle CAO$. Therefore, triangle $DAC$ is similar to triangle $AOC$. Consequently, $DC : AC = AC : OC$, or $2OC : AC = AC : OC$, from which $\frac{AC}{OC} = \sqrt{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_198acf192511df7a04ddg-46.jpg?height=520&width=886&top_left_y=0&top_left_x=583)

## Answer

$\sqrt{2}$.

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