Area of a square on top of a quadrilateral 
In quadrilateral $ABCD, \angle ABC + \angle DCB = 90^\circ$ and $ADEF$ is a square constructed on side $AD$ in the exterior of the quadrilateral $ABCD.$ If $BC = 10 $ cm, $AC = 9$ cm and $BD = 8$ cm, then the area of the square $ADEF$ lies between
(a) $70$ and $80$
(b) $60$ and $70$
(c) $50$ and $60$
(d) $40$ and $50$

I drew a rough sketch using Geogebra. Looks like the area of square should be between $40$ and $50.$
But I am not able to come with a proof. Any hints on how to proceed


Extend $\small CD$ to meet $\small AB$ at $\small E$.

Since $\small \measuredangle B+\measuredangle C=90^\circ$, we can see $\small \angle BEC=90^\circ.$
Now from Pythagorean theorem,
$\small BC^2=EB^2+EC^2\tag1$
$\small AD^2=EA^2+ED^2\tag2$
Adding both,
$\small \begin{align} BC^2+AD^2&=(EB^2+ED^2)+(EC^2+EA^2)\\ &=BD^2+AC^2\end{align}$
Now substituting,
$\small 10^2+AD^2=8^2+9^2\implies AD^2=\dots$
And $\small AD^2$ means ...
