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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.

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  • $\begingroup$ Any alternative approach is always welcomed. $\endgroup$
    – AbVk1718
    Aug 11, 2021 at 13:36

2 Answers 2

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enter image description here

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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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.

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  • $\begingroup$ Actually I mistyped it. It is actually BRAHMAGUPTA FORMULA, which can be used to find the area of any cyclic quadrilateral, given all its side lengths. You can find its proof here : en.wikipedia.org/wiki/Brahmagupta%27s_formula $\endgroup$
    – AbVk1718
    Aug 11, 2021 at 13:00
  • $\begingroup$ So, actually $AD$ and $BE$ are same segments? $\endgroup$
    – AbVk1718
    Aug 11, 2021 at 13:00
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    $\begingroup$ No , DE and DA are same segments . I had made a small mistake in writing the expression earlier. Now its fixed . And this could be easily visualized by another answer that contains the figure of this whole setup. and the answer from that answer is also the same answer , i.e. $2\sqrt(6)$ $\endgroup$
    – Smit
    Aug 11, 2021 at 15:31

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