By quick angle chasing note that if $$\angle DBC = \alpha $$then $\angle DAB = 2 \alpha$ and $\angle ADB = 90 + \alpha ; \angle ABD = 90 - 3 \alpha$ . So $\angle BDC = 45 - \alpha $ and $ \angle DCB = 135 $ . Let $BC \cap AD = \{F\} .$ Note that $FDC$ is right-angled in $F$ and with one angle of 45 . Evaluate $\sin \alpha $ and also $\cos \alpha$ and find $AB$ and $AD$ in terms of $BC = x$

Hi @above ... I am having a doubt here ... Can you please show me how $\angle$ $ADB$ comes out to be as $90 + \alpha$ ??

If $\angle ADB=x, \angle ABD=y,$ then $x-y=4a$ and $x+y=180^\circ -2a$ and by addition we get, $x=90^\circ +a.$

Ok ... thanks for helping me out @above ...

incorrect