过$A$点作$AF\bot CD$交$CD$的延长线于$F$点，如图，
$\because AE\bot BC$，$AF\bot CF$，
$\therefore \angle AEC=\angle CFA=90^{\circ}$，
而$\angle C=90^{\circ}$，
$\therefore$四边形$AECF$为矩形，
$\therefore \angle 2+\angle 3=90^{\circ}$，
又$\because \angle BAD=90^{\circ}$，
$\therefore \angle 1=\angle 3$，
在$\triangle ABE$和$\triangle ADF$中，
$\left\{\begin{array}{l}{∠1=∠3}\\{∠AEB=∠AFD}\\{AB=AD}\end{array}\right.$，
$\therefore \triangle ABE$≌$\triangle ADF\left(AAS\right)$，
$\therefore AE=AF=4$，$S_{\triangle ABE}=S_{\triangle ADF}$，
$\therefore$四边形$AECF$是边长为$4$的正方形，
$\therefore S_{四边形ABCD}=S_{正方形AECF}=4^{2}=16$，
故答案为：$16$.

过$A$点作$AF\bot CD$交$CD$的延长线于$F$点，如图，
$\because AE\bot BC$，$AF\bot CF$，
$\therefore \angle AEC=\angle CFA=90^{\circ}$，
而$\angle C=90^{\circ}$，
$\therefore$四边形$AECF$为矩形，
$\therefore \angle 2+\angle 3=90^{\circ}$，
又$\because \angle BAD=90^{\circ}$，
$\therefore \angle 1=\angle 3$，
在$\triangle ABE$和$\triangle ADF$中，
$\left\{\begin{array}{l}{∠1=∠3}\\{∠AEB=∠AFD}\\{AB=AD}\end{array}\right.$，
$\therefore \triangle ABE$≌$\triangle ADF\left(AAS\right)$，
$\therefore AE=AF=4$，$S_{\triangle ABE}=S_{\triangle ADF}$，
$\therefore$四边形$AECF$是边长为$4$的正方形，
$\therefore S_{四边形ABCD}=S_{正方形AECF}=4^{2}=16$，
故答案为：$16$.