Please Help Fast - Trigonometry

Consider the following diagram:

What are the lengths of AE and CE?

$\begin{array}{|rcll|} \hline CD^2 + 3^2 &=& 9^2 \\ CD^2 &=& 81 - 9 \\ CD^2 &=& 72 \\ CD &=& \sqrt{72} \\ \hline \end{array} \begin{array}{|rcll|} \hline BD^2 + 3^2 &=& 9^2 \\ BD^2 &=& 81 - 9 \\ BD^2 &=& 72 \\ BD &=& \sqrt{72} \\ \hline \end{array} \begin{array}{|rcll|} \hline CB &=& CD+BD \\ CB &=& \sqrt{72}+\sqrt{72} \\ CB &=& 2\sqrt{72} \\ CB^2 &=& 4*72 \\ \mathbf{CB^2} &=& \mathbf{288} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline AE^2+CE^2 &=& 9^2 \\ \mathbf{CE^2} &=& \mathbf{81-AE^2} \\ \hline \end{array} \begin{array}{|rcll|} \hline (9+AE)^2+CE^2 &=& CB^2 \\ (9+AE)^2+81-AE^2&=& 288 \\ 81+18AE+AE^2+81-AE^2&=& 288 \\ 81+18AE+81&=& 288 \\ 18AE&=& 288-2*81 \\ 18AE&=& 126 \\ \mathbf{AE}&=& \mathbf{7} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \mathbf{CE^2} &=& \mathbf{81-AE^2} \\ CE^2 &=& 81-7^2 \\ CE^2 &=& 81-49 \\ CE^2 &=& 32 \\ CE^2 &=& 16*2 \\ \mathbf{CE} &=& \mathbf{4*\sqrt{2}} \\ \hline \end{array}$

BC = 2[sqrt(9² - 3²)] = 2√72

CE = (3 * BC) / 9 = 4√2

AE = sqrt(AC² - CE²) = 7