In the following figure, $AD = AB$. Also $\angle DAB = \angle DCB = \angle AEC = 90^\circ$ and $AE = 5$. Find the area of quadrilateral $ABCD$. 
In the following figure, $AD = AB$. Also $\angle DAB = \angle DCB = \angle AEC = 90^\circ$ and $AE = 5$. Find the area of quadrilateral $ABCD$.

What I Tried: Here is the figure :-

I could conclude that $ABCD$ is cyclic, but I really could not use that property in a useful way till now. I thought about Ptolemy's Theorem but I am not sure since I don't know enough lengths.
Instead, Pythagoras Theorem gives the required values, assuming $AD = AB = x$ . But I am still stuck as that is not enough for finding the side-lengths of the quadrilateral, only then I can find it's area. The red angles came to be $45^\circ$, and I am stuck right here. I can't seem to use something with the cyclic quadrilaterals.
Can anyone help me? Thank You.
 A: (Fill in the gaps as needed. If you're stuck, show your work and explain where you're stuck.)
Hint: There is a degree of freedom in the problem.
Hence conclude that the area is $5\times 5 = 25$ by considering a special case (and assuming the area is an invariant).
This motivates how I thought of the next step.
Hint: Rotate $\Delta AED$ $90^\circ$ about $A$. What do we get?
(Note: There is slight work involved in proving this.)
Hence, conclude that the area is indeed 25.

A: Some brute force here and not at all elegant w.r.t Calvin Lin's answer..
Join $AC$ and hence see  by cyclic quad.$$\angle ACE=\angle ACB=45^{o}$$ hence $$EC=5,AC=5\sqrt{2}$$ thus $$2x^2={(\sqrt{x^2-25}+5)}^2+BC^2$$ Now apply cosine rule in $\Delta ABC$ to get $$\cos 45=\frac{x^2-50-BC^2}{10\sqrt{2}BC}$$
Solving you can do the rest easily...
A: Take $\angle BDC=\theta $, so $DE = xcos(\theta+45), xsin(\theta+45)=5, EC = x\sqrt{2}cos\theta-xcos(\theta+45), DC = x\sqrt{2}cos\theta, \\
BC = x\sqrt{2}sin\theta$

$\Delta ADB+\Delta BDC = \Delta ADE+ \square AECB$
$\Delta ADB+\Delta BDC = \dfrac{x^2}{2} + x^2sin\theta cos\theta = \dfrac{x^2}{2}(sin\theta + cos\theta )^2 $
$\Delta ADE+ \square AECB = x^2sin\theta cos\theta - \dfrac{x^2}{\sqrt{2}}sin\theta cos(\theta +45)+  \dfrac{5}{\sqrt{2}}xcos\theta  $
Equating both we get $x = \dfrac{5\sqrt{2}}{(sin\theta + cos\theta)}$
Using this in $\dfrac{x^2}{2}(sin\theta + cos\theta )^2 $  we get area $=25$
