6. A circle passes through vertices $A$ and $C$ of triangle $ABC$ and intersects its sides $AB$ and $BC$ at points $K$ and $T$ respectively, such that $AK: KB = 3: 2$ and $BT: TC = 1: 2$. Find $AC$, if $KT = \sqrt{6}$.
Answer: $3 \sqrt{5}$.

Solution. Let $B K=2 x, B T=y$; then $A K=3 x, C T=2 y$. By the theorem of two secants $B K \cdot B A=B T \cdot B C$, from which $2 x \cdot 5 x=y \cdot 3 y, y=x \sqrt{\frac{10}{3}}$. Triangles $A B C$ and $T B K$ are similar by two sides and the angle between them ( $B A: B T=B C: B K, \angle B-$ is common), and the similarity coefficient is $\frac{B C}{B K}=\frac{3 y}{2 x}=\frac{3}{2} \sqrt{\frac{10}{3}}=\sqrt{\frac{15}{2}}$. Therefore, $A C=K T \cdot \sqrt{\frac{15}{2}}=3 \sqrt{5}$.