Chords $A B$ and $C D$ intersect at point $E$ inside the circle. Let $M$ be an internal point of segment $B E$.

The tangent at point $E$ to the circumcircle of triangle $D E M$, intersects lines $B C$ and $A C$ at points $F$ and $G$ respectively. Let ${ }^{A M} /{ }_{A B}=t$. Find ${ }^{B G} /{ }^{E F}$.
Triangle $C E F$ is similar to triangle $A M D$, and triangle $C G E$ is similar to triangle $B D M$.

## Solution

Let point $E$ lie between points $G$ and $F$. Since $\angle C E F = \angle D E G = \angle E M D$, and $\angle E C F = \angle M A D$, triangles $C E F$ and $A M D$ are similar. Therefore,

$E F: D M = C E: A M$. On the other hand, since $\angle E C G = \angle M B D$, and $\angle C G E = \angle C E F - \angle G C E = \angle E M D - \angle M B D = \angle B D M$, triangles $C G E$ and $B D M$ are also similar. Consequently, $E G: D M = C E: M B$. Therefore, $E G / E F = A M / M B = t / 1-t$

![](https://cdn.mathpix.com/cropped/2024_05_06_cd0469c76668f3ff0930g-04.jpg?height=554&width=1200&top_left_y=692&top_left_x=429)

## Answer

$t / 1-t$.