Karel drew a triangle $A D E$. On side $A D$, he marked points $B$ and $C$ such that:

- point $C$ is the midpoint of side $A D$,
- triangles $A B E$ and $B C E$ have the same area,
- triangle $B D E$ is isosceles with base $B E$,
- the perimeter of triangle $A B E$ is $6 \mathrm{~cm}$ greater than the length of the broken line $E B C$,
- the perimeter of triangle $A C E$ is $2 \mathrm{~cm}$ greater than the perimeter of triangle $C D E$,

Calculate the perimeter of triangle $A D E$.

(E. Semerádová)
Triangles $A B E$ and $B C E$ have the same area and together form triangle $A C E$. Therefore, point $B$ is the midpoint of segment $A C$, or segments $A B$ and $B C$ are congruent. Triangles $A B E$ and the broken line $E B C$ have a common side $E B$, and $|A B|=|B C|$. The given difference thus corresponds to side $A E$, i.e., $|A E|=6 \text{ cm}$.

Point $C$ is the midpoint of segment $A D$, so segments $A C$ and $C D$ are congruent. Triangles $A C E$ and $C D E$ have a common side $C E$ and $|A C|=|C D|$. The difference in perimeters thus corresponds to the difference in the lengths of sides $A E$ and $D E$, i.e., $|A E|=|D E|+2 \text{ cm}$ or $|D E|=6-2=4 \text{ cm}$.

Triangle $B D E$ is isosceles with base $B E$, so $|D B|=|D E|=4 \text{ cm}$. Segments $A B$ and $B C$ are congruent, and segments $A C$ and $C D$ are congruent, so $|A B|=\frac{1}{4}|A D|$ or $|B D|=\frac{3}{4}|A D|$. From this, we get $|A D|=\frac{4}{3}|B D|=\frac{16}{3} \text{ cm}$.

The perimeter of triangle $A D E$ is

$$
|A D|+|D E|+|E A|=\frac{16}{3}+4+6=\frac{46}{3} \approx 15.33(\text{ cm})
$$

Evaluation. 3 points for auxiliary observations and partial results; 3 points for calculation and quality of commentary.

Note. The figure looks, to scale, like this:

![](https://cdn.mathpix.com/cropped/2024_04_17_f624dbe68821a0ba7fa9g-2.jpg?height=500&width=623&top_left_y=1937&top_left_x=725)