Problem 1969

Solution:

Extend sides $A B$ and $C D$ to intersect at point $F$. The angle conditions yield $\triangle B E C \sim \triangle A F D$, so $\angle A F D=90^{\circ}$. Therefore, since $\angle B F C$ and $\angle B E C$ are both right angles, quadrilateral $E B F C$ is cyclic and
$$\angle E F C=\angle E B C=90^{\circ}-\angle E C B=90^{\circ}-\angle E D F$$
implying that $E F \perp A D$.
Since $A F D$ is a right triangle, we have $\left(\frac{F A}{F D}\right)^{2}=\frac{A E}{E D}=\frac{1}{9}$, so $\frac{F A}{F D}=\frac{1}{3}$. Therefore $\frac{E B}{E C}=\frac{1}{3}$. Since the area of $C E D$ is 27 times more than the area of $A E B, E D=9 \cdot E A$, and $E C=3 \cdot E B$, we get that $\angle D E C=\angle A E B=45^{\circ}$. Since $B E C F$ is cyclic, we obtain $\angle F B C=\angle F C B=45^{\circ}$, so $F B=F C$.
Since $B C=6$, we get $F B=F C=3 \sqrt{2}$. From $\triangle E A B \sim \triangle E F C$ we find $A B=\frac{1}{3} F C=\sqrt{2}$, so $F A=4 \sqrt{2}$. Similarly, $F D=12 \sqrt{2}$. It follows that $A D^{2}=F A^{2}+F D^{2}=320$.