Area of Cyclic Quadrilateral 
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*Let $ABCD$ be a quadrilateral inscribed in a circle with diameter $AC$, and let $E$ be the foot of perpendicular from $D$ onto $AB$. If $AD=DC$ and the area of quadrilateral $ABCD$ is $24$, find $DE$.

Here's what I did:
Assuming the radius of the circle to be $r$, $AD=DC=a$, $AB=b$ and $BC=c$, I found out that:
• $a^2=2r^2$, and $b^2+ c^2 = 4r^2$,  using $Pythagoras$
•Next   $ $ $[ABCD]= \frac {a²}{2} + \frac {bc}{2} $.
Substituting the values, I got $b+c=4\sqrt6$.
•Finally, we have $$[ABCD] = \frac {(b+c)(b+c)(2a+b-c)(2a+c-b)}{16}  \\... using \  Brahmagupta \ Formula$$
Substituting values, I got $bc=24$ $\Rightarrow$ $ b=c=2\sqrt6$.
I am stuck here.
What should I do next?
Thanks.
 A: 
If $r$ is the radius of the circle, by Ptolemy Theorem,
$AC \cdot BD = AB \cdot CD + BC \cdot AD$
$2r \cdot BD = (AB + BC) \cdot AD \tag1$
As $AD = DC$, $\angle ABD = \angle CBD = 45^ \circ$
So, $\angle AOD = 90^ \circ \implies AD = r \sqrt2$
Plugging into $(1)$,
$BD = \cfrac{1}{\sqrt2} (AB + BC) \tag2$
Area of quadrilateral is,
$\cfrac{1}{2} (AB \cdot BD \sin 45^ \circ + BC \cdot BD \sin 45^ \circ) = 24$
$(AB+BC) \cdot BD = 48 \sqrt2 \tag3$
From $(2)$ and $(3)$, we easily see that $BD = 4 \sqrt3$ and then $DE = BD \sin 45^0$
A: Well I think you have the answer right there. if the you have that $b=c=2\sqrt6$ .then you have that the ABCD is actually a square . if you try and plot this scenario on figure it becomes pretty clear .
And even if you don't want to do that than just plug the values of b and c in the above relationship of
$ 24 = \dfrac{a^2}{2} + \dfrac{bc}{2} $
and you will get $a=b=c=2\sqrt6$ . and you have the square inscribed in the circle which would imply that A point coincides with E point and
$DE = DA = a = 2\sqrt6 $.
Thanks for the knowledge about the Brahmagupta formula . I didn't know about it before. Could you provide the derivation of it . I tried to find it but couldn't get it anywhere.  Hope this helped. sorry for any mishap in the typing. I am new to MSE.
