Is there a pure geometric solution to this problem? $ABCD$ is a square and there is a point $E$ such that $\angle EAB = 15^{\circ}$ and $\angle EBA = 15^{\circ}$. Show that $\triangle EDC$ is an equilateral triangle. 

Now there is a proof by contradiction to this problem. I was wondering if there is a pure geometric solution too (no trigonometry)?
 A: The best way to show that $\triangle CDE$ is equilateral is to construct an equilateral triangle $\triangle CDE'$ and argue that $\angle ABE'=\angle BAE'=15^\circ$. One can easily see this by looking at the isosceles triangles $\triangle ADE'$ and $\triangle BCE'$.  
It is then obvious that $E$ coincides with $E'$.

Here's another proof which is not "backward":

Construct $F$ such that $\angle BCF=\angle BCF=15^\circ$. It follows that $\triangle BEF$ is equilateral since $BE=BF$ and $\angle EBF=60^\circ$. 
Also $CF\perp EB$ as $\angle FCB+\angle EBC=90^\circ$. That means the line $CF$ is the perpendicular bisector of $BE$. Thus $CE=CD$.
A: Here is a geometric solution provided by Dr.Shailesh Shirali:
Draw a copy of triangle EAB with DA as base. That is, locate point F inside the square such that triangle FDA is congruent to triangle EAB.
Then angle FAD = angle EAB = 15 deg, so angle FAE = 60 deg. Also, FA = EA. Hence triangle FAE is equilateral, and angle AFE = 60 deg, and FA = FD = FE.
Therefore F is equidistant from A, D, E and so is the circumcentre of triangle DAE.
Therefore by the circle theorems, angle AFE = 2 angle ADE. 
But angle AFE = 60 deg. Hence angle ADE = 30 deg, and it follows that angle CDE = 60 deg. So triangle CDE is equilateral.

A: Reflect triangle $AED$ in $AE$ to obtain $AED'$.  Then $\angle BAD'$ is $60^{\circ}$.  Since $AB=AD=AD'$ this implies that $D'$ is straight below $E$ and therefore $\angle AED=\angle AED'=75^{\circ}$.  Then $\angle CDE=75^{\circ}-15^{\circ}=60^{\circ}$.
A: Pirsquare
Let $DC=a~$ and $EC=b$. Using cosine and sine theorems in triangles $EDC$ and $EBC$
, respectively we get that,$~a=2bcosx~$and $~\frac{sin(15+x)}{a}=\frac{sin75}{b}~$.These equations yield $\frac{sin(15+x)}{cosx}=2cos15~$. Finally, $tanx=2-tan15=\sqrt3$. So, $x=60~$.
A: Not sure it's what you mean by "proof by contradiction", anyway:
Don't assume you know the angles $EAB$ and $EBA$, but assume instead that the triangle $DCE$ is equilateral.
Then triangles $DEA, CEB$ are isoscele, thus $CEB=CBE=DEA=DAE$. But $EDA=ECB=90°-60°=30°$. Thus $DAE=CBE=75°$. Then $EAB=EBA=15°$.
Since there is only one position of $E$ on the mediatrix of $DC$ such that angle $BEA$ takes on a given value, your proposition is proved.
