Triangles $A B C$ and $G A C$ are similar.

## Solution

Let $C G=t, \angle C A G=\alpha$. Then $B C=3 t$, and since $C E$ is the height, $\angle B=\angle A C E$. Since trapezoid $A E F C$ is inscribed in a circle, it is isosceles, so $\angle A C E=\alpha$. Therefore, $\angle B=\alpha$. Consequently, right triangles $A B C$ and $G A C$ are similar by two angles. Thus,

$C G: A C=A C: B C$, which means $A C^{2}=C G \cdot B C$, or $12=3 t^{2}$, from which $t=2$. $\operatorname{tg} \alpha=C G / A C=\frac{1}{\sqrt{3}}$, so $\alpha=30^{\circ}$.

Therefore, $A G=2 C G=4$. By the tangent-secant theorem, $G F=C G^{2} / G A=1$.

![](https://cdn.mathpix.com/cropped/2024_05_06_4c124ad32a76c9faf91dg-14.jpg?height=654&width=368&top_left_y=1748&top_left_x=845)

## Answer

1.
[Similar auxiliary triangles $]$ [ Right triangle with an angle at ]

On the cathetus $A C$ of the right triangle $A B C$, a circle is constructed with $A C$ as its diameter. It intersects side $A B$ at point $E$. A point $G$ is taken on side $B C$ such that segment $A G$ intersects the circle at point $F$, and segments $E F$ and $A C$ are parallel, $B G=2 C G$ and $A C=2 \sqrt{3}$. Find $G F$.