6. A circle passes through the vertices $L$ and $M$ of triangle $F L M$ and intersects its sides $F L$ and $F M$ at points $A$ and $H$, respectively, different from the vertices of the triangle. The ratio of the area of triangle $F L M$ to the area of triangle $A F H$ is $\frac{49}{9}$.

a) Find the ratio $L M: A H$.

b) Suppose it is additionally known that the ratio of the areas of triangles $A F M$ and $F H L$ is $\frac{1}{4}$. Find the ratio $A L: M H$.
Answer: a) $L M: A H=7: 3$, b) $A L: M H=11$.

Solution. a) By the theorem of two secants $F L \cdot F A=F M \cdot F H$. Therefore, triangles $F L M$ and $A F H$ are similar by two sides and the angle between them ($F L: F H=F M: F A, \angle F$ - common). The areas of similar figures are in the ratio of the square of the similarity coefficient, so the similarity coefficient is $\frac{7}{3}$. Therefore, $L M: A H=7: 3$.

b) From the similarity proven in the first part, we also get that $F L=\frac{7}{3} F H, F M=\frac{7}{3} F A$. By the theorem of the ratio of areas of triangles with a common angle $\frac{S_{A F M}}{S_{F H L}}=\frac{F A}{F L} \cdot \frac{F M}{F H}$, from which $\frac{1}{4}=\frac{F A}{\frac{7}{3} F H} \cdot \frac{\frac{7}{3} F A}{F H}$. Therefore, $\frac{F A}{F H}=\frac{1}{2}$. Let $C B=3 x$, then $F H=6 x$, $F L=\frac{7}{3} F H=14 x$, $F M=\frac{7}{3} F A=7 x$, $H M=F M-F H=x$, $A L=F L-F A=11 x$, so $A L: M H=11$.