Area of an equilateral triangle divided by three lines An equilateral triangle is divided by three straight lines into seven regions whose areas are shown in the image below. Find the area of the triangle.

How to solve this problem ?
 A: By symmetry, the inner triangle is equilateral. Then the triangle marked with area $4$ is similar to one with area $28$ (on the smaller side of a cevian), so the similarity ratio is $7$. 
We can then apply Menelaus to one of the triangles composed of the inner equilateral triangle, two $20$ area triangles, and a small $4$ area triangle. The transversal is a cevian. If we let the smaller part of a side be $1$ arbitrary unit and the larger be $x$, we get 
$$\frac{x}{1}\frac{1+x}{1}\frac{1}{6} = 1.$$
We can now solve for $x$ and find the ratio that the cevian cuts the side into, which gives us the area ratio of the two triangles that the cevian cuts the triangle into, and hence the area of the small triangle.
A: Let the big triangle be $ABC$, the central $DEF$ and the remaining points be $P,Q,R$, so that we have lines $ADEP$, $BEFQ$, and $CFDR$. Let $x$ be the area of $\triangle DEF$.
Then $$AE:AP = \triangle ABE : \triangle ABP = (4+20):(4+20+4)=6:7$$
and
$$ AQ:AC = \triangle ABQ:\triangle ABC = (20+4+20+x):(4+20+4+20+4+20+x)=(44+x):(72+x).$$
Therefore
$$ (6\cdot(44+x)):(7\cdot(72+x))=\triangle AEQ:\triangle APC = (20+x):(20+4+20+x)$$
This should give you a quadratic equation in $x$ with only one positive solution.
A: 
Let the area of the equilateral triangle be $A$ and split the quadrilateral with area 20 into 2 triangles with area $x$ and $20-x$ respectively.
We shall use the basic property of triangle,
$$A:B=a:b$$
By symmetry, we get
$$
\frac{28}{44+A}\stackrel{(1)}{=}  \frac{4}{20-x} \stackrel{(2)}{=}\frac{4+x}{20+A}
$$
Simplifying (1) yields
$$
4 A=400-28 x \cdots (3)
$$
Simplifying (2) yields $$
x^{2}-16 x=4 A \cdots (4)$$
Combining (3) and (4) yields $$
\begin{array}{l}
x^{2}+12 x-400=0 \\
x=2(11- \sqrt{21})
\end{array}
$$
Putting back into (3) yields $$
\begin{aligned}
A &=14 \sqrt{21}-54
\end{aligned}
$$
