B2. Given a triangle $A B C$ with area 40. On side $A B$ lies point $D$ such that $|B D|=3 \cdot|A D|$. On side $B C$ lies point $E$ such that $|C E|=3 \cdot|B E|$. On side $C A$ lies point $F$ such that $|A F|=3 \cdot|C F|$.

Determine the area of triangle $D E F$.

![](https://cdn.mathpix.com/cropped/2024_04_17_8d46f403ab663a3b3700g-1.jpg?height=303&width=397&top_left_y=1542&top_left_x=1429)
B2. First, we will determine the area of triangle $B D E$. For this, we use triangle $B C D$. From the data in the problem, we deduce that $|B D|=\frac{3}{4} \cdot|A B|$ and $|B E|=\frac{1}{4} \cdot|B C|$.

When we compare the base $A B$ of triangle $A B C$ with the base $B D$ of triangle $B C D$, we see that $|B D|$ is a factor $\frac{3}{4}$ times as large as $|A B|$, while the triangles have the same height relative to these bases. The area of triangle $B C D$ is

![](https://cdn.mathpix.com/cropped/2024_04_17_351e9d9fc3725f0ab9f4g-1.jpg?height=365&width=500&top_left_y=1528&top_left_x=1349)
thus $\frac{3}{4} \cdot 40=30$.

Now we apply the same trick to triangles $B C D$ and $B D E$. The base $B E$ of triangle $B D E$ is $\frac{1}{4}$ times as large as the base $B C$ of triangle $B C D$. The heights are the same relative to these bases. The area of triangle $B D E$ is thus $\frac{1}{4} \cdot 30=\frac{15}{2}$.

In a similar manner, we can deduce that triangles $C E F$ and $A D F$ have an area of $\frac{15}{2}$. The area of triangle $D E F$ is therefore $40-3 \cdot \frac{15}{2}=\frac{35}{2}$.