<s>[INST] The diagram shows an equilateral triangle $ADE$ inside a square $ABCD$ . What is the value of $\frac{[\Delta ADE]}{[\Delta DEC]}$ . 
The diagram shows an equilateral triangle $ADE$ inside a square $ABCD$ . What is the value of $\frac{[\Delta ADE]}{[\Delta DEC]}$ .

What I Tried: Here is the diagram :-

You can see I marked the angles as required. Now let $AB = x$ . We then have :-
$$[\Delta ADE] = \frac{\sqrt{3}}{4}x^2$$
Now from here :- https://www.quora.com/What-is-the-ratio-of-sides-of-a-30-75-75-angle-triangle-without-sine-rule , I could understand and show that :- $$ED : DC : CE = \bigg(\frac{\sqrt{3} + 1}{2} : \frac{\sqrt{3} + 1}{2} : 1\bigg)$$
So let $EC = k$ , $CD = DE =  \frac{(\sqrt{3} + 1)k}{2}$ .
From here :- $$x = \frac{(\sqrt{3} + 1)k}{2}$$
$$\rightarrow k = EC = \frac{2x}{(\sqrt{3} + 1)}$$
Now, we can find area by Heron's Formula. We have :- $$s = x + \frac{x}{(\sqrt{3} + 1)}$$
$$\rightarrow s = \frac{x\sqrt{3} + 2x}{(\sqrt{3} + 1)}$$
So :- $[\Delta DEC] = \sqrt{s(s-a)(s-b)(s-c)}$
$$\rightarrow \sqrt{\Bigg(\frac{(x\sqrt{3} + 2x)}{(\sqrt{3} + 1)}\Bigg)\Bigg(\frac{x}{(\sqrt{3} + 1)}\Bigg)\Bigg(\frac{x}{(\sqrt{3} + 1)}\Bigg)\Bigg(\frac{(x\sqrt{3})}{(\sqrt{3} + 1)}\Bigg)}$$
This looks like really a complicated expression, and I really am not going to attempt to simplify this. So can anyone give me a different solution?
Thank You.
 [/INST] 
By Area of triangle using trigonometry
Area (DEC) $ = \frac12 (CD)(DE) \sin (30^\circ) = \frac 14 x^2$
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