In the following figure, $AD = AB$. Also $\angle DAB = \angle DCB = \angle AEC = 90^\circ$ and $AE = 5$. Find the area of quadrilateral $ABCD$.
What I Tried: Here is the figure :-
I could conclude that $ABCD$ is cyclic, but I really could not use that property in a useful way till now. I thought about Ptolemy's Theorem but I am not sure since I don't know enough lengths.
Instead, Pythagoras Theorem gives the required values, assuming $AD = AB = x$ . But I am still stuck as that is not enough for finding the side-lengths of the quadrilateral, only then I can find it's area. The red angles came to be $45^\circ$, and I am stuck right here. I can't seem to use something with the cyclic quadrilaterals.
Can anyone help me? Thank You.
Best Answer
(Fill in the gaps as needed. If you're stuck, show your work and explain where you're stuck.)
Hint: There is a degree of freedom in the problem.
Hence conclude that the area is $5\times 5 = 25$ by considering a special case (and assuming the area is an invariant).
This motivates how I thought of the next step.
Hint: Rotate $\Delta AED$ $90^\circ$ about $A$. What do we get?
(Note: There is slight work involved in proving this.)
Hence, conclude that the area is indeed 25.