3. Quadrilateral $A B C D$ is inscribed in a circle. A tangent line $\ell$ is drawn to this circle at point $C$. Circle $\omega$ passes through points $A$ and $B$ and is tangent to line $\ell$ at point $P$. Line $P B$ intersects segment $C D$ at point $Q$. Find the ratio $\frac{B C}{C Q}$, given that $B$ is the point of tangency of circle $\omega$.
Answer: 1.

![](https://cdn.mathpix.com/cropped/2024_05_06_627b343ad954e1e3a31bg-01.jpg?height=448&width=688&top_left_y=2083&top_left_x=701)

Solution. The angle between the tangent $BD$ and the chord $AB$ of the circle $\omega$ is equal to the inscribed angle that subtends $AB$, so $\angle APB = \angle ABD = \angle ACD$. Therefore, the quadrilateral $APCQ$ is cyclic, from which $\angle CQB = \angle CAP$. Applying the tangent-secant theorem twice more, we also get the equalities $\angle BCP = \angle BAC$ and $\angle BPC = \angle BAP$. Then,

$$
\angle CBQ = 180^{\circ} - \angle CBP = \angle BCP + \angle BPC = \angle BAC + \angle BAP = \angle CAP = \angle CQB
$$

Thus, triangle $BCQ$ is isosceles, from which $BC = CQ$.