There is a $\triangle ABC$ which $\angle C=90$, and $\bar{AB}=36$ On the circumcircle of $\triangle ABC$, there is $\overarc{BC}$ which does not include point $A$. D is on $\overarc{BC}$. It satisfies $2\times\angle CAD = \angle BAD $ $E: \bar{AD}\cap\bar{BC} $ $ \bar{AE}=20 $ Find $ \bar{BD}^2 $