jmc

First of all, a sketch might be useful. Since we have an isosceles triangle on our hands, let's drop a median/altitude/bisector from $B$ as well: We might be able to create some usable right triangles if we draw a perpendicular segment from $D$ to $AC$: Thanks to $AA$ similarity, we see that $△DFC\sim △BEC.$ We see that $CD:CB=DF:BE=CF:CE.$ As for $CD:CB,$ we know that $CD:DB=4:5$ by the Angle Bisector Theorem. Since $CB=CD+DB,$ it follows that $CD:CB=DF:BE=CF:CE=4:9.$ That means $DF=BE\cdot \left(\frac{4}{9}\right),$ and $CF=CE\cdot \left(\frac{4}{9}\right).$

Since $CE$ is half of $AC,$ we have that $CE=2$ and $CF=\frac{8}{9}.$ Then, $AF=AC-FC=4-\frac{8}{9}=\frac{28}{9}.$

We apply the Pythagorean Theorem to find that $A{D}^2=D{F}^2+A{F}^2.$ We just found $AF,$ and as for $DF,$ we have $DF=BE\cdot \left(\frac{4}{9}\right).$ Squaring both sides, we have $D{F}^2=B{E}^2\cdot \left(\frac{16}{81}\right).$ We know that $B{E}^2=B{C}^2-C{E}^2=5^2-2^2=21.$ Therefore, $D{F}^2=21\cdot \left(\frac{16}{81}\right).$

Going back to the expression for $A{D}^2,$ we now have $$\begin{array}{rl}A{D}^2& =D{F}^2+A{F}^2\\ & =21\cdot \left(\frac{16}{81}\right)+{\left(\frac{28}{9}\right)}^2\\ & =\frac{336}{81}+\frac{784}{81}=\boxed{\frac{1120}{81}}.\end{array}$$