As $B=2C \Rightarrow A=\pi-B-C=\pi-3C$ then
$ 0
and $A=\pi-3C$ so, $0
So, $0< C< \frac{\pi}{3} $
If $∠BAD=x , ∠DAC=x$ and if $∠ADB=y, ∠ADC=\pi-y$
If $AB=c$, $BC=a$, $CA=b \Rightarrow BD=a-c$
Applying the law of sines on triangle $ACD$, $\frac{c}{\sin x} = \frac{b}{\sin (\pi- y)}\Rightarrow\frac{c}{\sin x} = \frac{b}{\sin y}$
Applying sine law on triangle ABD, $\frac{a-c}{\sin x} = \frac{c}{\sin y}$
Upon division, $\frac{a-c}{c}=\frac{c}{b}$
$\frac{\sin\ A - \sin C}{\sin C}=\frac{\sin C}{\sin B}$ (applying $a=2R\sin A$ etc.)
$\frac{\sin\ (\pi - 3C) - \sin C}{\sin C}=\frac{\sin C}{\sin 2C}$
$\frac{\sin\ 3C - \sin C}{\sin C}=\frac{\sin C}{\sin 2C}$
$(2\sin C \cos2C)\sin2C = (\sin C)^2$
applying $\sin2x-\sin2y=2\sin(x-y)\cos(x+y)$
$\sin C(2 \cos2C\sin2C - \sin C)=0$
$\sin C(\sin 4C - \sin C)=0$
$\sin 4C=\sin C \text{ as } 0\sin C>0$
$4C=n\pi+(-1)^nC$ where n is any integer,
If $n$ is even $(=2m)$,say, $4C=2m\pi+C \Rightarrow C=\frac{2m\pi}{3}$,
As $0< C< \frac{\pi}{3} $, so there is no solution if $n$ is even.
If $n$ is odd $(=2m+1)$, say, $4C=(2m+1)\pi-C\Rightarrow C=\frac{(2m+1)\pi}{5}$
$C=\frac{\pi}{5}\ as\ 0< C< \frac{\pi}{3} $
$A=\pi-3C=\frac{2\pi}{5}=72^\circ$