Let $AB=a, BC=DA=b, CD=c$. I assume $AB \ge CD$. Then $a+c=2b$. We know $BE = \frac{a-c}{2}$. Then it suffices to show $b^2-\left( \frac{a-c}{2} \right)^2 = ac$ or $4b^2-a^2+2ac-c^2 = 4ac$ and using $a+c=2b$ it reduces to $(a+c)^2 - a^2-c^2 = 2ac$, true.

same as mine solution!

Nice geometry. My solution: Let the points of tangency of the incircle be $P,Q,R,S$ respectively with $AB,BC,CD,DA$. By symmetry, $AP=PB$ and $DR=RC$. If $O$ is the incentre, clearly $P,O,R$ are collinear (which can be seen by joining $OS$ or $OQ$). Let us join $OS$. Also, note that $CE=PR=2*r$ and $PR$ is perpendicular to $AB$ and $CD$. In triangle $AOD$, $\angle OAD + \angle ODA = \frac{1}{2}* (\angle PAD + \angle RDA) = 90^\circ$. Thus, $\angle AOD=90^\circ$. Since $OS$ is perpendicular to $AD$, $AS*SD=OS^2=r^2$. Thus, $AB*CD=(2AP)*(2DR)=4*AS*SD=4r^2=(2r)^2=CE^2$.