## Task 28/63

Given is a triangle $A B C$. The side $a=BC$ is extended beyond $C$ by $u_{a}$ to $C^{\prime}$, the side $b=CA$ is extended beyond $A$ by $v_{b}$ to $A^{\prime}$, and the side $c=AB$ is extended beyond $B$ by $w_{c}$ to $B^{\prime}$.

What is the area $F^{\prime}$ of the triangle $A^{\prime} B^{\prime} C^{\prime}$, measured in areas $F$ of the triangle $A B C$? (Figure)

![](https://cdn.mathpix.com/cropped/2024_06_06_1d2be4837fb839c2b31dg-077.jpg?height=457&width=625&top_left_y=331&top_left_x=1184)
Let $F_{1}$ be the area and $h_{a}^{\prime}$ the height of triangle $C C^{\prime} A^{\prime}$, $F_{2}$ the area and $h_{b}^{\prime}$ the height of triangle $A A^{\prime} B^{\prime}$, and $F_{3}$ the area and $h_{c}^{\prime}$ the height of triangle $B B^{\prime} C^{\prime}$. Then the following equations hold (see figure):

$$
F_{1}=\frac{1}{2} u a h_{a}^{\prime} \quad ; \quad F_{2}=\frac{1}{2} v b h_{b}^{\prime} \quad ; \quad F_{3}=\frac{1}{2} w c h_{c}^{\prime}
$$

![](https://cdn.mathpix.com/cropped/2024_06_06_1d2be4837fb839c2b31dg-077.jpg?height=698&width=817&top_left_y=1116&top_left_x=625)

From the intercept theorem, it follows that

$$
\begin{aligned}
h_{a}^{\prime}: h_{a}=\frac{b+v b}{b}=1+v & ; & h_{a}^{\prime}=(1+v) h_{a} \\
h_{b}^{\prime}: h_{b}=\frac{c+w c}{c}=1+w & ; & h_{b}^{\prime}=(1+w) h_{b} \\
h_{c}^{\prime}: h_{c}=\frac{a+u a}{a}=1+u & ; & h_{c}^{\prime}=(1+u) h_{c}
\end{aligned}
$$

Thus, we get

$$
F_{1}=\frac{1}{2} u a(1+v) h_{a} \quad ; \quad F_{2}=\frac{1}{2} v b(1+w) h_{b} \quad ; \quad F_{3}=\frac{1}{2} w c(1+u) h_{c}
$$

If $\frac{1}{2} a h_{a}=\frac{1}{2} b h_{b}=\frac{1}{2} c h_{c}=F$, it follows that

$$
F_{1}=F u(1+v) \quad ; \quad F_{2}=F v(1+w) \quad ; \quad F_{3}=F w(1+u)
$$

For $F^{\prime}$, we have $F^{\prime}=F+F_{1}+F_{2}+F_{3}$, thus

$$
F^{\prime}=F+F u(1+v)+F v(1+w)+F w(1+u)=F(1+u+v+w+u v+v w+w u)
$$