Given chords $AB, CD$ of a circle intersect at point $E$ inside the circle. Let $M$ be an interior point of segment $EB$. The circle passing through points $D, E, M$ is tangent to lines $BC$ and $AC$ at points $F$ and $G$, respectively. If $H$ is such that $\frac{AM}{AB}=t$, express $\frac{EG}{EF}$ in terms of $t$.

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A given circle's chords $AB, CD$ intersect at point $E$ inside the circle. Let $M$ be an interior point of segment $EB$. The circle passing through points $D, E, M$ is tangent to lines $BC$ and $AC$ at points $F$ and $G$, respectively. If $H$ is such that $\frac{AM}{AB}=t$, express $\frac{EG}{EF}$ in terms of $t$.
The point where $M D$ intersects the circle for the second time is denoted by $X$. Furthermore, let $\angle A D C = \gamma$, $\angle C D X = \alpha$, and $\angle X D B = \beta$. Since it also belongs to the arc $A C$, $\angle A B C = \gamma$. The line $F G$ is the tangent to the circumcircle of triangle $D E M$ at point $E$, and in this small circle, the angle $\angle F E M$ is the angle subtended by the chord $E M$ and the tangent, so $\angle F E M = \angle E D M = \alpha$. For its vertical angle: $\angle G E A = \alpha$. From now on, we will refer to the circumcircle of $A B C D$ as "the circle". Choosing the radius of this circle to be $\frac{1}{2}$, and applying the general sine theorem for any three points $L_{1}, L_{2}, L_{3}$ on the circle:

$$
L_{1} L_{3} = \left(\sin \angle L_{1} L_{2} L_{3}\right) \cdot 2 R = \sin \angle L_{1} L_{2} L_{3}
$$

Applying the sine theorem in triangle $A E G$:

$$
\frac{E G}{\sin \angle G A E} = \frac{A E}{\sin \angle A G E}
$$

Here, we make two observations: since $\angle G A E + \angle E A C = 180^{\circ}$, it follows that $\sin \angle G A E = \sin \angle E A C$. Furthermore, since the angles subtended by $C B$ are inscribed angles, $\alpha + \beta = \angle C D B = \angle C A B$ and $\angle C A B$ is the exterior angle of triangle $A E G$, so $\alpha + \beta = \angle C A B = \angle A G E + \angle A E G$, thus $\angle A E G = \alpha$ and $\angle A G E = \beta = \angle X D B$.

From $(*)$ and $(* *)$:

$$
\frac{E G}{C B} = \frac{A E}{X B}
$$

Notice that in triangle $E F B$, the angle at $E$ is $\alpha$, the angle at $B$ is $\gamma$, so the angle at $F$ is $180^{\circ} - \alpha - \gamma$, and $\sin \angle E F B = \sin (\alpha + \gamma) = A X$. Applying the sine theorem in triangle $E F B$: $\frac{E F}{\sin \gamma} = \frac{E B}{\sin (\alpha + \gamma)}$, thus from $(*)$:

$$
\frac{E F}{A C} = \frac{E B}{A X}
$$

Dividing (1) by (2):

$$
\frac{E G}{E F} = \frac{A E \cdot C B \cdot A X}{E B \cdot A C \cdot X B}
$$

**Lemma**: For a convex cyclic quadrilateral $A C B D$, the intersection of diagonals $A B$ and $C D$ is $E$. Then:

$$
C B \cdot A E \cdot B D = A C \cdot E B \cdot A D
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_a356e29600ccc80186cdg-1.jpg?height=569&width=631&top_left_y=1556&top_left_x=736)

**Proof**: In triangle $A D E$, by the sine theorem:

$$
\frac{A E}{\sin \angle A D E} = \frac{A D}{\sin \angle A E D},
$$

thus from $(*)$:

$$
\frac{A E}{A C} = \frac{A D}{\sin \angle A E D}.
$$

In triangle $E B C$ similarly:

$$
\begin{gathered}
\frac{E B}{\sin \angle B C E} = \frac{C B}{\sin \angle C E B} = \frac{C B}{\sin \angle A E D}, \\
\frac{E B}{B D} = \frac{C B}{\sin \angle A E D}.
\end{gathered}
$$

Dividing equation $(L_{2})$ by equation $(L_{1})$ and rearranging, we obtain the statement to be proven.

Applying the lemma to quadrilaterals $A C B D$ and $A X B D$:

$$
\begin{aligned}
C B \cdot A E \cdot B D & = A C \cdot E B \cdot A D \\
X B \cdot A M \cdot B D & = A X \cdot M B \cdot A D
\end{aligned}
$$

Dividing the second equation by the first:

$$
\frac{A M}{M B} = \frac{A E \cdot C B \cdot A X}{X B \cdot E B \cdot A C}
$$

Comparing this with (3):

$$
\frac{E G}{E F} = \frac{A M}{M B} = \frac{A M}{A B - A M} = \frac{t}{1 - t}
$$

**Remark**. After reading the problem and drawing the diagram, I immediately had the impression that there are so many equal angles here that it is inconceivable that the solution could not be "extracted" in some way. Unfortunately, not finding an elegant solution, I had to do this. In my competition solution, I applied the sine theorem to 8 triangles, and by multiplying these "chains" together, many terms canceled out, and I obtained the statement to be proven. The lemma used in the solution here is suggested by János Pataki, which makes the solution much more comprehensible (although the background still involves the same sine theorems).