<s>[INST] Feyman's Triangle? How do you find the area of the inner triangle if the outside triangle is equilateral If triangle $ABC$ is equilateral,$BD/BC=1/3, CE/CA=1/3,$ and $ AF/AB=1/3$. What is the ratio of the area of triangle? I have problems analyzing this triangle I tried to use phythagorean, heron's formula, the formula $A$=$\frac{ab\sin(\theta)}{2}$ but I still can't figure the inside area. I've also done some research they said it's $1:7$? But what I want is to understand how they got $ 1:7$.

 [/INST] There are many proofs of this result (see here for a generalized version), but perhaps the easiest is using the theorem of Menelaus, along with some area ratios.

We use this figure. Consider collinear points $A, K, E$ on the sides of $\triangle DBG$. Menelaus's Theorem tells us:
$$\begin{align}
\frac{DA}{AB} \frac{BK}{KG} \frac{GE}{ED} &= 1 \\
\frac{3}{1} \frac{BK}{KG} \frac{2}{1} &= 1\\
\implies \frac{BK}{KG} &= \frac{1}{6}\\
\implies \frac{KG}{BG} &= \frac{6}{7}
\end{align}$$
Thus 
$$\begin{align}
[\triangle KAG] &= \frac{KG}{BG} [\triangle BAG] \\
&= \frac{6}{7} \frac{BA}{DA}[\triangle DAG]\\
&= \frac{6}{7} \frac{1}{3} [\triangle DAG]\\
&= \frac{6}{21} [\triangle DAG]
\end{align}$$
By symmetry, we have that 
$$ [\triangle KAG] = [\triangle LDA] = [\triangle MGD] = \frac{6}{21}[\triangle DAG] $$
Thus, 
$$
\begin{align}
[\triangle LKM] &= [\triangle DAG] - ([\triangle KAG] + [\triangle LDA] + [\triangle MGD])\\
&= ( 1 - \frac{6}{21} - \frac{6}{21} - \frac{6}{21})[\triangle DAG])\\
&= \frac{1}{7}[\triangle DAG]
\end{align}$$
As desired.
</s>