Try this on for size:
Let z = length(AB).
It is clear through the use of similar triangles that DEC is isosceles.
Choose F to bisect DC and erect the perpendicular, which intersects AB at G.
Since ABCD is a square, FG is perpendicular to AB.
Since DEC is isosceles, E is on FG.
Draw the circular arc of radius z, centered at C, from B to D.
Draw the circular arc of radius z, centered at D, from A to C.
These arcs obviously intersect at some point E', and moreover it is clear that DE'C is equilateral.
We will show that E' = E.
Since DE'C is equilateral, it is isosceles. Thus E' is on FG.
Since DE'C is equilateral, angle(E'CB) = 30.
Since triangle E'CB is isosceles, we have
angle(CE'B) = angle(CBE') = 75. Thus angle(E'BA) = 15.
Recall that we were given angle(EBA) = 15.
Consider the two triangles E'GB and EGB.
By the angle-side-angle rule, these triangles are congruent.
Then since E' and E are both on segment FG, we have
E' = E QED