11. (IND $\left.3^{\prime}\right)^{\mathrm{IMO}}$ Given a circle with two chords $A B, C D$ that meet at $E$, let $M$ be a point of chord $A B$ other than $E$. Draw the circle through $D, E$, and $M$. The tangent line to the circle $D E M$ at $E$ meets the lines $B C, A C$ at $F, G$, respectively. Given $\frac{A M}{A B}=\lambda$, find $\frac{G E}{E F}$.
11. Assume $\mathcal{B}(A, E, M, B)$. Since $A, B, C, D$ lie on a circle, we have $\angle G C E = \angle M B D$ and $\angle M A D = \angle F C E$. Since $F D$ is tangent to the circle around $\triangle E M D$ at $E$, we have $\angle M D E = \angle F E B = \angle A E G$. Consequently, $\angle C E F = 180^{\circ} - \angle C E A - \angle F E B = 180^{\circ} - \angle M E D - \angle M D E = \angle E M D$ and $\angle C E G = 180^{\circ} - \angle C E F = 180^{\circ} - \angle E M D = \angle D M B$. It follows that $\triangle C E F \sim \triangle A M D$ and $\triangle C E G \sim \triangle B M D$. From the first similarity we obtain $C E \cdot M D = A M \cdot E F$, and from the second we obtain $C E \cdot M D = B M \cdot E G$. Hence 
$$ \begin{gathered} A M \cdot E F = B M \cdot E G \Longrightarrow \\ \frac{G E}{E F} = \frac{A M}{B M} = \frac{\lambda}{1-\lambda} . \end{gathered} $$
![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-543.jpg?height=371&width=353&top_left_y=1023&top_left_x=894)
If $\mathcal{B}(A, M, E, B)$, interchanging the roles of $A$ and $B$ we similarly obtain $\frac{G E}{E F} = \frac{\lambda}{1-\lambda}$.