4. The side of an equilateral triangle $A B C$ is 4 cm long. The orthogonal projections of the midpoint $D$ of side $A B$ onto sides $B C$ and $A C$ are denoted by $E$ and $F$. Calculate the area of triangle $D E F$.
$\mathrm{I} / 4$.

1. way Clearly, $|D A|=|D B|=\frac{|A B|}{2}=2$. Triangle $D B E$ has angles equal to $60^{\circ}, 90^{\circ}$ and $30^{\circ}$, so it is equal to half of an equilateral triangle, thus $|B E|=\frac{|B D|}{2}=1$. Similarly, for triangle $A D F$, it is congruent to triangle $B D E$. The length of side $D E$ or $D F$ is equal to $\frac{\sqrt{3}}{2} \cdot|B D|=\sqrt{3}$. The area of each triangle is thus equal to $\frac{\sqrt{3}}{2}$.

Triangle $E C F$ has sides of length $|C F|=|C A|-|A F|=3$ and $|C E|=|C B|-|E B|=3$, so it is an isosceles triangle with an angle of $60^{\circ}$ at the vertex, i.e., equilateral. The area of this triangle is thus equal to $\frac{3^{2} \cdot \sqrt{3}}{4}$.

![](https://cdn.mathpix.com/cropped/2024_06_07_24964e3bc413fb5af60fg-07.jpg?height=486&width=445&top_left_y=1479&top_left_x=1388)

The area of triangle $D E F$ can thus be calculated by subtracting the areas of triangles $D B E, A D F$ and $E C F$ from the area of triangle $A B C$. We get that the desired area is equal to

$$
\frac{4^{2} \cdot \sqrt{3}}{4}-\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}-\frac{3^{2} \cdot \sqrt{3}}{4}=\frac{3 \sqrt{3}}{4}
$$

![](https://cdn.mathpix.com/cropped/2024_06_07_24964e3bc413fb5af60fg-07.jpg?height=57&width=1633&top_left_y=2239&top_left_x=220)

![](https://cdn.mathpix.com/cropped/2024_06_07_24964e3bc413fb5af60fg-07.jpg?height=60&width=1633&top_left_y=2289&top_left_x=220)

Calculation that $|D E|=\sqrt{3}$ or $|D F|=\sqrt{3} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$

![](https://cdn.mathpix.com/cropped/2024_06_07_24964e3bc413fb5af60fg-07.jpg?height=54&width=1633&top_left_y=2389&top_left_x=220)

![](https://cdn.mathpix.com/cropped/2024_06_07_24964e3bc413fb5af60fg-07.jpg?height=52&width=1633&top_left_y=2441&top_left_x=220)

Calculation of the area of triangle $C E F$............................................................................................................

Calculation of the area of triangle $D E F$.......................................................................

2. way Clearly, $|D A|=|D B|=\frac{|A B|}{2}=2$. Triangles $A F D$ and $B E D$ have angles equal to $60^{\circ}, 90^{\circ}$ and $30^{\circ}$ and match in the length of the hypotenuse, so they are congruent. This means $|D E|=|D F|$, so triangle $D E F$ is isosceles. Let $G$ be the intersection of segments $C D$ and $E F$. Since $\angle F D G=\angle G D E=60^{\circ}$, $D G$ is the height of triangle $D E F$. Therefore, triangles $D G E$ and $D G F$ are also congruent and have angles equal to $60^{\circ}, 90^{\circ}$ and $30^{\circ}$.

Triangles $A D C, A F D$ and $D G F$ are thus similar. From the similarity of $A D C$ and $A F D$ it follows that $\frac{|A D|}{|A C|}=\frac{|A F|}{|A D|}$, from which we get $|A F|=1$. Similarly, $\frac{|A D|}{|D C|}=\frac{|A F|}{|F D|}$, from which we get $|F D|=\sqrt{3}$.

![](https://cdn.mathpix.com/cropped/2024_06_07_24964e3bc413fb5af60fg-08.jpg?height=474&width=443&top_left_y=237&top_left_x=1389)

The similarity of triangles $A D C$ and $D G F$ gives us $\frac{|A D|}{|A C|}=\frac{|D G|}{|D F|}$, so $|D G|=\frac{\sqrt{3}}{2}$. Similarly, from $\frac{|A D|}{|D C|}=\frac{|D G|}{|G F|}$ we get $|G F|=\frac{3}{2}$.

The area of triangle $D E F$ is equal to the sum of the areas of triangles $F G D$ and $D G E$, so it is equal to

$$
|D G| \cdot|F G|=\frac{3 \sqrt{3}}{4}
$$

![](https://cdn.mathpix.com/cropped/2024_06_07_24964e3bc413fb5af60fg-08.jpg?height=369&width=1665&top_left_y=1069&top_left_x=204)

3. way Clearly, $|D A|=|D B|=\frac{|A B|}{2}=2$. Triangles $A F D$ and $B E D$ have angles equal to $60^{\circ}, 90^{\circ}$ and $30^{\circ}$ and match in the length of the hypotenuse, so they are congruent. Thus, $|D E|=|D F|=\sqrt{3}$. Let $G$ be the midpoint of segment $E F$. Since $E F \| A B$, triangles $F G D$ and $E G D$ are congruent right triangles and $\angle D F G=\angle D E G=30^{\circ}$. Therefore,

$$
p_{D E F}=2 p_{D F G}=2 \cdot \frac{1}{2} \frac{\sqrt{3}^{2} \cdot \sqrt{3}}{4}=\frac{3 \sqrt{3}}{4}
$$

![](https://cdn.mathpix.com/cropped/2024_06_07_24964e3bc413fb5af60fg-08.jpg?height=486&width=449&top_left_y=1573&top_left_x=1386)
Determination that $|A D|=2$ or $|B D|=2$ ..... 1 point
Calculation that $|B E|=1$ or $|A F|=1$ ..... 1 point
Calculation that $|D E|=\sqrt{3}$ or $|D F|=\sqrt{3}$ ..... 1 point
Determination that segment $C D$ is perpendicular to segment $E F$ ..... 1 point
Calculation that $|D G|=\frac{\sqrt{3}}{2}$ ..... 1 point
Calculation that $|F G|=\frac{3}{2}$. ..... 1 point
Calculation of the area of triangle $D E F$ 1 point