8. On the sides $B C, C A, A B$ of an equilateral triangle $A B C$ with side length 7, points $A_{1}, B_{1}, C_{1}$ are taken respectively. It is known that $A C_{1}=B A_{1}=C B_{1}=3$. Find the ratio of the area of triangle $A B C$ to the area of the triangle formed by the lines $A A_{1}, B B_{1}, C C_{1}$.
Solution:

Triangles $A B A_{1}, B C B_{1}, C A C_{1}$ are equal by sides and the angle between them, triangles $A D C_{1}, B E A_{1}, C F B_{1}$ are equal by side and angles. Triangle $A D C_{1}$ is similar to $A B A_{1}$ by two angles, triangle $A B C$ is similar to $D E F$. Let $S=S_{A B C}, S_{1}=S_{A B A_{1}}, S_{2}=S_{A C_{1} D}, S_{0}=S_{D E F} \cdot$ Then

$$
\begin{aligned}
& \qquad A A_{1}=\sqrt{A B^{2}+B A_{1}^{2}-2 A B \cdot B C_{1} \cos 60^{\circ}}=\sqrt{49+9-21}=\sqrt{37} \\
& \frac{S_{2}}{S_{1}}=\left(\frac{A C_{1}}{A A_{1}}\right)^{2}=\frac{9}{37}, \frac{S_{1}}{S}=\frac{B A_{1}}{B C}=\frac{3}{7} \\
& S_{0}=S-3 S_{1}+3 S_{2}=S-3 \cdot \frac{3}{7} S+3 \cdot \frac{9}{37} \cdot \frac{3}{7} S=\frac{S}{37}, \frac{S}{S_{0}}=37 \\
& \text { answer: } 37 \text {. }
\end{aligned}
$$

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