6. A circle passes through the vertices $A$ and $N$ of triangle $A C N$ and intersects its sides $A C$ and $C N$ at points $B$ and $K$, respectively, different from the vertices of the triangle. The ratio of the area of triangle $B C K$ to the area of triangle $A C N$ is $\frac{1}{4}$.

a) Find the ratio $A N: B K$.

b) Suppose it is additionally known that the ratio of the areas of triangles $B C N$ and $A C K$ is $\frac{9}{16}$. Find the ratio $N K: A B$.
Answer: a) $A N: B K=2$, b) $N K: A B=2: 5$.

Solution. a) By the theorem of two secants $C K \cdot C N=C B \cdot C A$. Therefore, triangles $A C N$ and $K C B$ are similar by two sides and the angle between them ($A C: K C=C N: C B, \angle C$ - common). The areas of similar figures are in the ratio of the square of the similarity coefficient, so the similarity coefficient is 2. Therefore, $A N: B K=2$.

b) From the similarity proved in the first part, we also get that $A C=2 C K, C N=2 B C$. By the theorem of the ratio of areas of triangles with a common angle $\frac{S_{B C N}}{S_{A C K}}=\frac{C B}{C A} \cdot \frac{C N}{C K}$, from which $\frac{9}{16}=\frac{C B}{2 C K} \cdot \frac{2 B C}{C K}$. Therefore, $\frac{C B}{C K}=\frac{3}{4} . \quad$ Let $\quad C B=3 x, \quad$ then $\quad C K=4 x, \quad A C=2 C K=8 x, \quad C N=2 B C=6 x, \quad A B=A C-B C=5 x$, $N K=C N-C K=2 x$, so $N K: A B=2: 5$.