In an equilateral triangle $A B C$, an equilateral triangle $D E F$ is inscribed, see the figure. The vertices $D, E$, and $F$ lie on the sides $A B, B C$, and $A C$ such that the sides of triangle $D E F$ are perpendicular to the sides of triangle $A B C$. Furthermore, the segment $D G$ is a median in triangle $D E F$, and point $H$ is the intersection of the lines $D G$ and $B C$.

Determine the ratio of the areas of triangles $H G C$ and $B E D$.

(E. Patáková)

![](https://cdn.mathpix.com/cropped/2024_04_17_60411719b4f3e7daffe2g-2.jpg?height=617&width=646&top_left_y=548&top_left_x=705)
From the symmetry of the entire figure (or from the given sentence), it follows that triangles $B E D$ and $C F E$ are congruent. We will therefore determine the ratio of the areas of triangles $H G C$ and $C F E$.

Triangle $D E F$ is equilateral, so its medians and altitudes coincide, and median $D G$ is thus perpendicular to segment $E F$. According to the problem, segments $E F$ and $A C$ are also perpendicular, so lines $A C$ and $D G$, as well as $F C$ and $G H$, are parallel. Since $D G$ is the median of triangle $D E F$, point $G$ lies at the midpoint of segment $E F$. Therefore, $G H$ is the midline of triangle $C F E$.

![](https://cdn.mathpix.com/cropped/2024_04_17_60411719b4f3e7daffe2g-2.jpg?height=617&width=642&top_left_y=1573&top_left_x=707)

Since triangles $H G C$ and $H G E$ share the side $H G$ and the heights corresponding to this side are the same (specifically, $|F G|=|G E|$), these triangles have the same area. Since triangles $H G E$ and $C F E$ are similar and the corresponding sides are in the ratio $1: 2$, the areas of these triangles are in the ratio $1: 4$. (Side $G H$ is half the length of side $F C$, and the height of triangle $H G E$ to side $G H$ is half the height of triangle $C F E$ to side $F C$; thus, the area of the first triangle is one-fourth the area of the second.)

The ratio of the areas of triangles $H G C$ and $B E D$ is $1: 4$.

Note. If we denote the length of segment $B E$ as $a$, we can express the areas of triangles $H G C$ and $B E D$ in terms of this quantity.

From the parallelism of lines $A C$ and $D H$, it follows that triangle $D B H$ is equilateral and triangle $B E D$ is half of it. Therefore, $|B D|=|B H|=2 a$ and the length of $D E$ can be calculated using the Pythagorean theorem in triangle $B E D$:

$$
|D E|=\sqrt{(2 a)^{2}-a^{2}}=a \sqrt{3} .
$$

Triangle $D E F$ is equilateral, so $|E F|=|D E|=a \sqrt{3}$. Point $G$ is at the midpoint of segment $E F$, so

$$
|G F|=\frac{a \sqrt{3}}{2}
$$

Triangles $B E D$ and $C F E$ are congruent, so $|C F|=|B E|=a$. Segment $G H$ is the midline of triangle $C F E$, so

$$
|G H|=\frac{a}{2} .
$$

Now we can express the areas of the triangles being studied:

$$
\begin{aligned}
S_{B E D} & =\frac{1}{2}|B E| \cdot|D E|=\frac{1}{2} a^{2} \sqrt{3} \\
S_{H G C} & =\frac{1}{2}|G H| \cdot|G F|=\frac{1}{8} a^{2} \sqrt{3}
\end{aligned}
$$

Evaluation. 1 point for identifying the parallelism of lines $A C$ and $D G$; 2 points for determining $|G H|=\frac{1}{2}|F C|$; 3 points for expressing the desired areas or their ratio.