6. A circle passes through the vertices $A$ and $K$ of triangle $A K T$ and intersects its sides $A T$ and $K T$ at points $C$ and $N$ respectively, such that $A C: C T=4: 1, T N: N K=1: 2$. Find $A K$, if $C N=\sqrt{10}$.
Answer: $5 \sqrt{6}$.

Solution. Let $C T=x, T N=y$; then $A C=4 x, K N=2 y$. By the theorem of two secants $T C \cdot T A=T N \cdot T K$, from which $x \cdot 5 x=y \cdot 3 y, y=x \sqrt{\frac{5}{3}}$. Triangles $A K T$ and $C N T$ are similar by two sides and the angle between them ( $A T: N T=K T: C T, \angle T-$ is common), and the similarity coefficient is $\frac{A T}{N T}=\frac{5 x}{y}=5 \sqrt{\frac{3}{5}}=\sqrt{15}$. Therefore, $A K=C N \cdot \sqrt{15}=5 \sqrt{6}$.