Let ABCD be a quadrilateral inscribed in a circle with diameter AC, an

Correct Option : D


First, we have that BC || ED , which is obvious from the information.
Next is that since AD=DC , BD is the angle bisector of ∠BAC which gives BED an isosceles triangle and DE=BE.
Now draw a perpendicular from C to DE and let it intersect at F .
We have that ΔAED≅ΔDFC from RHS , and this gives all the necessary piece of info in the picture, also noting that CFEB is a rectangle.
This total areas of the 2 triangles and the rectangle gives the area of the quadrilateral

From here :-
[ABCD] = 2 [ΔAED] + [CFEB]
→24 = x(x+y) + y(x+y)
→24 = (x+y)^2
→(x+y) = DE = √24

Hence DE^2 = 24 .

PS : This is not my solution, found while serching for solution on web, credit goes to the person, who has solved it, sharing as a learning for better of all.

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Can anybody explain how to proceed ?

andreagonzalez2k highly interesting approach. Thanks for sharing
v12345 very well and detailed explanation. Thanks for sharing. Need to ensure I remember subtend property

How area of BCD = 1/2*BC*X

thakurarun85
BC extended to drop a perpendicular from D = height BCD = x
Area of triangle BCD = 1/2 * base * height = 1/2 * BC * x

Thank you for this.

That first piece of information is key.

If from the same Line BA

A line segment ED and a line segment BC both “depart” or extend from that Line BA from the same Angle (90 degrees) ———-> then ED and BC will be parallel.

Hard question.