Geometry proof given diagram 
$ABCD$ forms a square. $CDE$ forms a triangle. Given $\measuredangle AED=15^{\circ}$ and $DE=CE$, prove $\triangle CDE$ is equilateral. 
The question is surprising hard, the problem is basically proving $\measuredangle BAE=\measuredangle AEB$. Can I have some hint. 
 A: Each possible length of $DE$ both is uniquely determined by and uniquely determines angle $\measuredangle CDE$. Likewise, $\measuredangle DEA$ is uniquely determined by and uniquely determines either one of the length $DE$ or $\measuredangle CDE$. This means that we can reverse the problem! Thus "Prove that if triangle $CDE$ is equilateral, $\measuredangle DEA=15^{\circ}$" is an equivalent problem. This is much easier. Proof:
1) Triangle $DEA$ is isoceles (since $DE=AD$)
2)$\measuredangle ADE=(90+60)^{\circ}=150^{\circ}$
3)$\measuredangle DEA=((180-150)/2)^{\circ}=15^{\circ}$
And we are done.
A: Ok ! Here we go ..
$ABCD$ is a square.
That means -> $AB=BC=CD=AD$
Now if you carefully notice, $CD$ is the base of the $\Delta CDE$
Given that $DC = DE$
We can conclude that $\Delta ADE$ is an isosceles triangle.
Therefore ,
$\angle AED = \angle EAD$
So, $\angle EAD = 15º$
Now, By Angle-Sum Property in $\Delta ADE$
$\angle AED + \angle EAD + \angle ADE = 180º $
=> $\angle ADE = 150º $
=> $90º + \angle CDE = 150º$
=> $\angle CDE = 60º$
Also you can notice that $\Delta DCE$ isosceles.
Therefore , 
$\angle DEC = \angle DCE$
By Angle-Sum Property in $\Delta DEC$ 
$\angle CDE + \angle DCE + \angle CED = 180º$
=> $60º + \angle DCE + \angle DCE = 180º$
=> $2\angle DCE = 120º$
=> $\angle DCE = 60º$
Therefore,
$\angle CDE = \angle DCE = \angle CED = 60º$
So $\Delta DCE$ is equilateral !
Hence Proved 
