ABC is an isosceles triangle -prove AD=BC ABC is an isosceles triangle having $\angle B=\angle C=2*\angle A$. If BD bisecting $\angle B$ meets AC in D,prove that AD=BC.
I know congruent triangles would help but am not able to figure out how to use them. ADB can not be congruent to CBD. I am trying to figure out which triangles might be congruent(triangles with AD as side and BC as side).
 A: See angles are $72,72,36$ of triangle ABC . So BD is bisector of B implies each angle=36. So $AD=BD ...(1)$isoceles triangle theorem now in triangle BDC angle $BDC=72$ but also $DCB =72$ so $BD=BC$..(2) isoceles triangle theorem thus from 1,2 $AD=BC$ thas all.
A: 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}
A: By construction we have:
$$
\angle DBA=\angle DCB=\frac{1}{2} \angle ABC=\frac{1}{2} \angle ACB=\angle CAB
$$
so, for the triangle $ADB$:
$$
\angle DAB= \angle CAB=\frac{1}{2} \angle ABC=\angle DBA \Rightarrow DA=DB
$$ 
and for the triangle $CDB$:
$$
\angle  CDB=180°-(\angle DCB+\angle DBC)=(180°-\angle DBC)-\angle DCB=
$$
$$
=(180°-\angle CAB)-\angle DCB=2\times \angle DCB-\angle DCB=\angle DCB \Rightarrow DB=CB
$$
So: $DA=DB$
