Problem 7. Let $OP$ be the diameter of circle $\Omega$, and $\omega$ be a circle centered at point $P$ with a radius smaller than that of $\Omega$. Circles $\Omega$ and $\omega$ intersect at points $C$ and $D$. Chord $OB$ of circle $\Omega$ intersects the second circle at point $A$. Find the length of segment $AB$, if $BD \cdot BC = 5$.
Answer: $\sqrt{5}$.

Solution. Let $N-$ be the second intersection point of line $O A$ with $\omega ; M$ - the second intersection point of line $D B$ with $\omega$.

Note that due to symmetry with respect to line $O P$, arcs $O C$ and $O D$ are equal. Therefore, $\angle A B C=$ $\angle D B A$. Let this angle be $\alpha$.

Then, from the cyclic quadrilateral $C B P D$, $\angle C P D=\angle C B D=2 \alpha$. Consequently, since $P-$ is the center of $\omega, \breve{D A}+\breve{A C}=\widetilde{D A C}=2 \alpha$. On the other hand, $\widetilde{D A}+\widetilde{M N}=2 \angle D B A=2 \alpha$. Subtracting the common arc $\widetilde{D A}$, we get that $\breve{A C}=\breve{M N}$, hence $\breve{A M}=\breve{C N}$.

Thus, $\angle C A B=\angle A D B$, and triangles $A B C$ and $D B A$ are similar by two angles, from which $A B / B C=B D / A B$, or $A B^{2}=B C \cdot B D=5, A B=\sqrt{5}$.