Simplest known solution to "World's Hardest Easy Geometry Problem"?

I've seen multiple solutions to this well-known problem (posed, among other sites, at http://thinkzone.wlonk.com/MathFun/Triangle.htm). None of the solutions are particularly simple. I was wondering if trigonometry (which the site I mentioned instructs you not to use) would yield an easier solution. Can anybody provide one?

EDIT: Multiple users have marked this as a duplicate question. The question they are referring to is far more general and probably far more difficult. All the answers that were given to it either lacked proofs or were incorrect. I'm looking for a specific, detailed, simple proof for this particular problem that takes advantage of trigonometry instead of elementary geometry.

Solution 1:

First, take note that $\triangle BCD$ is isosceles, and that $\angle BDC$ is 140 degrees.

Assuming arbitrarily sides $BD$ and $CD$ are of unit length, we can use the law of cosines with $\angle BDC$ to obtain that the length of $CB$ is $\approx 1.87939$.

Similarly, notice that triangle ABC is isosceles. $AC$ must then be congruent to $CB$, and have the same measure of $\approx 1.87939$.

Using law of cosines with $AC$, $CB$, and $\angle ACB$, we determine that $AB$ has measure $\approx 0.652705$.

To determine the length of $AD$, we use the law of cosines with $AB$, $BD$, and $\angle ABD$. We learn that $AD$ has measure $\approx 0.879385$.

Using the law of sines with $\angle BAE$, $\angle BEA$, and $AB$, we can determine that the length of $BE$ is $\approx 1.22668$.

Using the law of cosines with $BD$, $BE$, and $\angle DBE$, we can determine the length of $DE$, which is $\approx 0.446475$.

Finally, using the law of sines, we can calculate the angle measure of $\angle BED$ using all three sides of the involved triangle, $BE$, $DE$, and $BD$. This angle is 50 degrees. Since $\angle BEA$ was 30 degrees, we can finally conclude that angle $\angle DEA$ is 20 degrees!