## Task 4 - 180724

For six points $A, B, C, D, E, F$, the following is assumed:

$\triangle A B C$ is a right triangle with $B$ as the vertex of the right angle. $D$ is an (internal) point on the segment $A B, E$ is an (internal) point on the segment $B C, F$ is an (internal) point on the segment $D B$.

The triangles $A D C, D E C, D F E$ and $F B E$ are all equal in area. Furthermore, $F B=15 \mathrm{~cm}$ and $B E=20 \mathrm{~cm}$.

Determine the length of the segment $A D$ under these assumptions!
For the area $A_{1}$ of triangle $F B E$, according to the given information and the area formula for triangles,

$$
A_{1}=\frac{1}{2} \cdot 15 \cdot 20 \mathrm{~cm}^{2}=150 \mathrm{~cm}^{2}
$$

The area of triangle $D B E$ is given as $2 \cdot A_{1}$, so for $B D$ it follows that:

$$
\frac{1}{2} B D \cdot B E=300 \mathrm{~cm}^{2}, \text { i.e., } B D=30 \mathrm{~cm}
$$

The area of triangle $D B C$ is given as

$$
\frac{1}{2} B C \cdot B D=450 \mathrm{~cm}^{2}, \text { i.e., } B C=30 \mathrm{~cm}
$$

Similarly, for $A B$:

$$
\frac{1}{2} A B \cdot B C=600 \mathrm{~cm}^{2}, \text { i.e., } A B=40 \mathrm{~cm}
$$

and thus $A D=A B-B D=40 \mathrm{~cm}-30 \mathrm{~cm}=10 \mathrm{~cm}$. The length of segment $A D$ is $10 \mathrm{~cm}$.

Solutions of the 2nd Round 1978 adopted from [5]

### 4.20.3 3rd Round 1978, Class 7