4. In triangle $ABC$, the angle at vertex $A$ is twice the angle at vertex $C$. A tangent line $\ell$ to the circumcircle $\Omega$ of triangle $ABC$ is drawn through vertex $B$. The radius of the circle $\Omega$ is $5 \sqrt{2}$, and the distances from points $A$ and $C$ to the tangent line $\ell$ are in the ratio $2:5$.

a) Find the ratio of the distances from point $A$ to the lines $\ell$ and $BC$.

b) Find the distance from point $C$ to the line $\ell$ and the length of side $AB$.
Answer. a) $2: 3$; b) $A B=5 \sqrt{3}, d_{C}=\frac{75}{4 \sqrt{2}}$.

Solution. Let $\angle A C B=\gamma$, then $\angle B A C=2 \gamma$. By the generalized sine theorem, we find that $A B=2 R \sin \gamma, B C=2 R \sin 2 \gamma$. Note that $\gamma<90^{\circ}$ (otherwise the sum of the angles in the triangle would be greater than $180^{\circ}$).

By the theorem on the angle between a tangent and a chord, the angle between the tangent and $A B$ is $\gamma$, and the angle between the tangent and $B C$ is $2 \gamma$. Therefore, the distance $d_{A}$ from point $A$ to the tangent is $A B \sin \gamma=2 R \sin ^{2} \gamma$. Similarly, the distance $d_{C}$ from point $C$ to the tangent is $B C \sin 2 \gamma=2 R \sin ^{2} 2 \gamma$.

Thus, $\frac{5}{2}=\frac{d_{C}}{d_{A}}=4 \cos ^{2} \gamma$, from which $\cos ^{2} \gamma=\frac{5}{8}$. Then $\sin ^{2} \gamma=\frac{3}{8}, d_{A}=\frac{15}{2 \sqrt{2}}, d_{C}=\frac{75}{4 \sqrt{2}}$.

$A B=2 R \sin \gamma=2 \cdot 5 \sqrt{2} \cdot \frac{\sqrt{3}}{2 \sqrt{2}}=5 \sqrt{3}$.

The distance from point $A$ to line $B C$ is

$$
A B \cdot \sin \angle A B C=A B \sin 3 \gamma=A B\left(3 \sin \gamma-4 \sin ^{3} \gamma\right)=\frac{45}{4 \sqrt{2}}
$$

The ratio of the distances from point $A$ to lines $\ell$ and $B C$ is $\frac{15}{2 \sqrt{2}}: \frac{45}{4 \sqrt{2}}=\frac{2}{3}$.