Instead of looking for congruent triangles, you should instead find the angle of triangles themselves.
We know that :
\begin{equation} 2\times \angle A=\angle B =\angle C\end{equation}
Since $\angle A, \angle B , \angle C $ are parts of single triangle then :
\begin{equation} \angle A+\angle B +\angle C=\pi\end{equation}
solve for these angles we would find:
\begin{equation} \angle A= \frac{\pi}{5}, \angle B =\angle C=\frac{2\pi}{5}\end{equation}
$BD$ bisect $\angle B$ means that
\begin{equation} \angle CBD = \angle ABD =\frac{\pi}{5}\end{equation}
Which tells us that triangle $ABD$ is isosceles thus
\begin{equation} AD =BD\end{equation}
We can also solve for $\angle BDC$ to find that
\begin{equation}\angle BDC=\frac{2\pi}{5}\end{equation}
thus triangle $BDC$ is also isosceles and
\begin{equation}BD=BC\end{equation}
finally we proved that
\begin{equation}AD=BC\end{equation}