Let $CD,DA,AB,BC$ be denoted by $a,b,c,d,$ respectively. The condition $AD\cdot CE=DC\cdot AE$ implies that $DB$ is the bisector of $\angle ADC.$ Therefore, $\angle ADC=45^\circ,$ $\angle ABC=135^\circ,$ and $c=d.$ Using the law of cosines in $\triangle ABD,$ and $\triangle CBD,$ we see that $a$ and $b$ are the solutions of the quadratic $x^2-12x\cos 22.5^\circ+36-c^2=0,$ and therefore $ab=36-c^2.$ However, the area of the quadrilateral is given by ${1\over 2}\cdot (ab+cd)\cdot \sin 45^\circ={1\over 2}\cdot (36-c^2+c^2)\sin 45^\circ=9\sqrt{2}.$

I hope solution is true

Attachments: