In an acute triangle $ABC$, point $D$ is on the segment $AC$ such that $\overline{AD}=\overline{BC}$ and $\overline{AC}^2-\overline{AD}^2=\overline{AC}\cdot\overline{AD}$. The line that is parallel to the bisector of $\angle{ACB}$ and passes the point $D$ meets the segment $AB$ at point $E$. Prove, if $\overline{AE}=\overline{CD}$, $\angle{ADB}=3\angle{BAC}$.