Find $\triangle GCH$ 
$ABCD$ is a unit square and $CEF$ is an equilateral triangle with side length $ = 1$. If  $BCE$ is  a straight line and $AE$ intersects $CD$ and $CF$ at points $G$ and $H$ respectfully. Find the area of  $\triangle GCH$.


My attempt:
Its trivial to prove that $\triangle EGC \sim \triangle EAB$, it follows that $CG = \frac{1}{2}$. Now,  $$[GCE] = \frac{1}{2}(EC)(GC) = \frac{1}{4}$$
and, $$\frac{[GCH]}{[ECH]} = \frac{\frac{1}{2}(\sin30^\circ)(GC)(CH)}{\frac{1}{2}(\sin60^\circ)(EC)(CH)}= \frac{1}{2\sqrt 3}$$
Also, $$ [GCH] + [ECH] =[GCE] $$
From the above equations we get our answer. But I want to see a pure geometry solution to this problem.(I tried by dropping an altitude from $F$ and then using similar triangles.)
 A: One trigonometry-free approach uses the co-side theorem: if lines $AB$ and $CD$ intersect at $X$, then $AX : BX = S_{ACD} : S_{BCD}$. (Here, the ratio should really be a signed ratio and the areas $S_{ACD}, S_{BCD}$ should be signed areas. In this problem, we will ignore the signs, because we know the ordering of all points.)
From this, we can compute $GH : EH = S_{GCF} : S_{ECF}$. Thinking of $\triangle GCF$ as having base $CG = \frac12$ and height $\frac12$, we see that $S_{GCF} = \frac18$; meanwhile, we know in any number of ways that $S_{ECF} = \frac{\sqrt3}{4}$. Therefore $GH : EH = \frac18 : \frac{\sqrt3}{4} = 1 : 2\sqrt3$.
Equivalently, $GH$ is $\frac1{1+2\sqrt3}$ of $GE$. It follows that $S_{CGH}$ is $\frac1{1+2\sqrt3}$ of $S_{CGE}$. We know that $S_{CGE} = \frac14$; therefore $S_{CGH} = \frac1{4+8\sqrt3}$.
A: Yet another approach. Draw from $H$ the line perpendicular to $BE$, and let $K$ be its intersection with $BE$. You have that
$$\overline{CH} = 2\overline {CK}$$
and by Pythagorean Theorem
$$\overline{HK} = \sqrt 3 \ \overline {CK}.$$
Since $\triangle ABE \sim \triangle HKE$, we have
$$\overline{KE} = 2\sqrt 3 \ \overline{CK},$$
and, since $\overline{CK} + \overline{KE} = 1$ we obtain
$$\overline{CK} = \frac1{1+2\sqrt 3}.$$
Your result follows easily by computing the required area as
$$\mathcal [GCH] = \frac{\overline{CG}\cdot \overline{CK}}2.$$
