3. In a convex quadrilateral $A B C D$, diagonals $A C$ and $D B$ are perpendicular to sides $D C$ and $A B$ respectively. A perpendicular is drawn from point $B$ to side $A D$, intersecting $A C$ at point $O$. Find $A O$, if $A B=4, O C=6$.
Answer: 2. Solution. Since $\angle A B D=\angle A C D=90^{\circ}$, the quadrilateral is inscribed in a circle with diameter $A D$. Therefore, $\angle A C B=\angle A D B$ - as angles subtending the same arc. Since $B H$ is the altitude in the right $\triangle A B D$, then $\angle A D B=\angle A B H$. Then triangles $A B O$ and $A C B$ are similar by two angles, and the following ratios hold: $\frac{A B}{A O}=\frac{A C}{A B} \Leftrightarrow \frac{4}{A O}=\frac{A O+6}{4}$ $\Leftrightarrow A O^{2}+6 A O-16=0 \Leftrightarrow A O=-8$ or $A O=2$. Answer: $A O=2$.

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## Answer for variant 152: 9.

Answer for variant 153: 1.

Answer for variant 154: 12.