<s>[INST] Area of Cyclic Quadrilateral 
*

*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.
 [/INST] 
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 $DE = BD \sin 45^0$
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