<s>[INST] Geometry/Triangles Problem Let $A,B,C$ be a triangle's vertices (FIG.1), and $D$ is a point on $\overline {BC}$. It is given that $\angle B= 2 \angle C $,  $\overline{AB}=\overline{DC}$ and also that $\overline{AD}$ is the bisector of angle $A$.. We have to prove that angle $\angle BAC=72°$.
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$$\begin{align}\text{FIGURE 1} &\end{align}$$
 [/INST] As $B=2C \Rightarrow A=\pi-B-C=\pi-3C$ then
$$ 0<B<\pi \Rightarrow 0<2C<\pi \Rightarrow 0<C<\frac{\pi}{2}$$
and $A=\pi-3C$ so, $0<A<\pi \Rightarrow0<\pi - 3C <\pi \Rightarrow0< C< \frac{\pi}{3} $
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 $$\sin2x-\sin2y=2\sin(x-y)\cos(x+y)$$
$$\sin C(2 \cos2C\sin2C - \sin C)=0$$
$$\sin C(\sin 4C - \sin C)=0$$
$$\sin 4C=\sin C  \text{ as } 0<C<\pi=>\sin C>0$$
$4C=n\pi+(-1)^nC$ 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, $4C=(2m+1)\pi-C\Rightarrow C=\frac{(2m+1)\pi}{5}$
$$C=\frac{\pi}{5}\ as\ 0< C< \frac{\pi}{3} $$
$$A=\pi-3C=\frac{2\pi}{5}=72^\circ$$
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