HMMT November

Extend sides $AB$ and $CD$ to intersect at point $F$. The angle conditions yield $\triangle BEC \sim \triangle AFD$, so $\angle AFD=90^{\circ}$. Therefore, since $\angle BFC$ and $\angle BEC$ are both right angles, quadrilateral $EBCF$ is cyclic and $$\angle EFC=\angle BEC=90^{\circ}-\angle ECB=90^{\circ}-\angle EDF$$ implying that $EF \perp AD$. Since $AFD$ is a right triangle, we have $\left(\frac{FA}{FD}\right)^{2}=\frac{AE}{ED}=\frac{1}{9}$, so $\frac{FA}{FD}=\frac{1}{3}$. Therefore $\frac{EB}{EC}=\frac{1}{3}$. Since the area of $CED$ is 27 times more than the area of $AEB, ED=9 \cdot EA$, and $EC=3 \cdot EB$, we get that $\angle DEC=\angle AEB=45^{\circ}$. Since $BECF$ is cyclic, we obtain $\angle FBC=\angle FCB=45^{\circ}$, so $FB=FC$. Since $BC=6$, we get $FB=FC=3\sqrt{2}$. From $\triangle AEB \sim \triangle EFC$ we find $AB=\frac{1}{3}FC=\sqrt{2}$, so $FA=4\sqrt{2}$. Similarly, $FD=12\sqrt{2}$. It follows that $AD^{2}=FA^{2}+FD^{2}=320$.