Help Appreciated

Let's see what i can do..

So you need lengths AE and CE

Apparently Angle ADC is not a right angle?

First, we know that $DB =DC = \sqrt{9^2 - 3^2} = 6\sqrt{2}$, which means $CB = 12 \sqrt 2$

Also, note that $\triangle CEB \sim \triangle ADB$ by AAA. This is because both triangles have a right angle and share $\angle B$, which means that all 3 traingles have the same angles, and thus, are similar. 

Now, by similiar triangles, we have ${AB \over CB} = {AD \over CE}$. Substituting what we know, we have ${9 \over 12 \sqrt 2} = {3 \over CE}$, meaning $CE = \color{brown}\boxed{4 \sqrt2}$.

To solve for AE, we use the Pythagorean Theorem on $\triangle AEC$, and find that $AE = \sqrt{9^2 - {4\sqrt2}^2} = \color{brown}\boxed{7}$

Now that I think of it, here's another way of reaching the same answer. 

We already know that $CB = 12\sqrt2$ (look at the other solution to see how)

Now, let $CE = a$ and $AE = b$. We have $a^2 + b ^2 = 81$, and $a^2 + (b+9)^2 = ({12 \sqrt2})^2 = 288$. 

Subtracting the two equations gives us $18b + 81 = 207$, meaning $b = 7$. 

Plugging this into the first equation also gives us $a = 4 \sqrt 2$