A $k$ circle has a regular hexagon $A B C D E F$ inscribed in it. The intersection of lines $C D$ and $B F$ is point $G$, and the reflection of $G$ over $A B$ is $H$. Calculate the ratio of the lengths of the chords cut by lines $A G$ and $A H$ from $k$.
Since the hexagon is regular, the diagonal $B E$ is an axis of symmetry. The line $A F$ is the mirror image of $C D$ with respect to this axis. The intersection of the lines $F B$ and $C D$ is $G$, which is also the mirror image of $F$ with respect to $B$. The mirror image of $D$ with respect to $B$ is $D^{\prime}$, which lies on the line $F A$ and coincides with the mirror image of $G$ with respect to the axis $E B$.

![](https://cdn.mathpix.com/cropped/2024_05_02_24bba5604d460ab684b1g-1.jpg?height=1192&width=1049&top_left_y=301&top_left_x=533)

Due to the axial symmetry with respect to $A B$, and since $D^{\prime} D$ is also perpendicular to the line $A B$, the triangle $A D^{\prime} G$ is congruent to the triangle $A D H$.

Now let's start the calculations. Let the side of the hexagon be of unit length. Denote the intersection of the chord $A G$ with the circle by $X$, and the chord $A H$ by $Y$. We need to determine the ratio $A X : A Y$.

To do this, we will write the area of the triangles $A D H$ and $A G D$ in two different ways.

From the triangle $A D H$, we have $A D = 2$, and $A$ and $D$ are opposite points of the hexagon. From the congruence and symmetry,

$$
D H = D^{\prime} G = F D = \sqrt{3}, \quad \text{and} \quad A H = \sqrt{A D^{2} + D H^{2}} = \sqrt{7}
$$

The double area of the triangle $A D H$ is thus

$$
A D \cdot D H = A H \cdot D Y \quad \text{and} \quad D Y = \sqrt{D A^{2} - A Y^{2}}
$$

Substituting,

$$
2 \sqrt{3} = \sqrt{7} \cdot \sqrt{4 - A Y^{2}}
$$

From this,

$$
A Y = \frac{4}{\sqrt{7}}
$$

From the triangle $A G D$, we have $A D = 2$, $A G = \sqrt{7}$, and $G D = G C + C D = A D + C D = 3$. Similarly, writing the double areas,

$$
\begin{gathered}
A G \cdot D X = D G \cdot A C \quad \text{and} \quad D X = \sqrt{A D^{2} - A X^{2}} \\
\sqrt{7} \cdot \sqrt{4 - A X^{2}} = 3 \cdot \sqrt{3}
\end{gathered}
$$

From this, $A X = \frac{1}{\sqrt{7}}$. Thus, the desired ratio is $\quad A X : A Y = 1 : 4$.