6. A circle passes through the vertices $Q$ and $E$ of triangle $M Q E$ and intersects its sides $M Q$ and $M E$ at points $B$ and $D$, respectively, different from the vertices of the triangle. The ratio of the area of triangle $B D M$ to the area of triangle $M Q E$ is $\frac{9}{121}$.

a) Find the ratio $Q E: B D$.

b) Suppose it is additionally known that the ratio of the areas of triangles $B M E$ and $D Q M$ is 4. Find the ratio $B Q: D E$.
Answer: a) $Q E: B D=11: 3$, b) $B Q: D E=5: 19$.

Solution. a) By the theorem of two secants $M Q \cdot M B=M E \cdot M D$. Therefore, triangles $M Q E$ and $M D B$ are similar by two sides and the angle between them ( $M Q: M D=M E: M B, \angle M-$ is common). The areas of similar figures are in the ratio of the square of the similarity coefficient, so the similarity coefficient is $\frac{11}{3}$. Therefore, $Q E: B D=11: 3$.

b) From the similarity proven in the first part, we also get that $M Q=\frac{11}{3} M D, M E=\frac{11}{3} M B$. By the theorem of the ratio of areas of triangles with a common angle $\frac{S_{B M E}}{S_{D Q M}}=\frac{M B}{M Q} \cdot \frac{M E}{M D}$, from which $4=\frac{M B}{\frac{11}{3} M D} \cdot \frac{\frac{11}{3} M B}{M D}$. Therefore, $\frac{M B}{M D}=2$. Let $M D=3 x$, then $M B=6 x, \quad M Q=\frac{11}{3} M D=11 x, \quad M E=\frac{11}{3} M B=22 x, \quad D E=M E-M D=19 x$, $B Q=M Q-M B=5 x$, so $B Q: D E=5: 19$.