# 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.

• Any alternative approach is always welcomed. Aug 11, 2021 at 13:36

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 $$DE = BD \sin 45^0$$

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.

• 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 Aug 11, 2021 at 13:00
• So, actually $AD$ and $BE$ are same segments? Aug 11, 2021 at 13:00
• 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. $2\sqrt(6)$
– Smit
Aug 11, 2021 at 15:31