In a triangle $\angle A = 2\angle B$ iff $a^2 = b(b+c)$ Prove that in a triangle $ABC$, $\angle A = \angle 2B$, if and only if:
$$a^2 = b(b+c)$$
where $a, b, c$ are the sides opposite to $A, B, C$ respectively. 
I attacked the problem using the Law of Sines, and tried to prove that if $\angle A$ was indeed equal to $2\angle B$ then the above equation would hold true. Then we can prove the converse of this to complete the proof. 
From the Law of Sines, 
$$a = 2R\sin A = 2R\sin (2B) = 4R\sin B\cos B$$
$$b = 2R\sin B$$
$$c = 2R\sin C = 2R\sin(180 - 3B) = 2R\sin(3B) = 2R(\sin B\cos(2B) + \sin(2B)\cos B)$$
$$=2R(\sin B(1 - 2\sin^2 B) +2\sin B\cos^2 B) = 2R(\sin B -2\sin^3 B + 2\sin B(1 - \sin^2B))$$
$$=\boxed{2R(3\sin B - 4\sin^3 B)}$$
Now, 
$$\implies b(b+c) = 2R\sin B[2R\sin B + 2R(3\sin B - 4\sin^3 B)]$$
$$=4R^2\sin^2 B(1 + 3 - 4\sin^2 B)$$
$$=16R^2\sin^2 B\cos^2 B = a^2$$
Now, to prove the converse:
$$c = 2R\sin C = 2R\sin (180 - (A + B)) = 2R\sin(A+B) = 2R\sin A\cos B + 2R\sin B\cos A$$
$$a^2 = b(b+c)$$
$$\implies 4R^2\sin^2 A = 2R\sin B(2R\sin B + 2R\sin A\cos B + 2R\sin B\cos) $$
$$ = 4R^2\sin B(\sin B + \sin A\cos B + \sin B\cos A)$$
I have no idea how to proceed from here. I tried replacing $\sin A$ with $\sqrt{1 - \cos^2 B}$, but that doesn't yield any useful results.

$$a^2-b^2=bc\implies \sin^2A-\sin^2B=\sin B\sin C\text{ as }R\ne0$$
Now, $\displaystyle\sin^2A-\sin^2B=\sin(A+B)\sin(A-B)=\sin(\pi-C)\sin(A-B)=\sin C\sin(A-B)\  \  \   \  (1)$
$$\implies \sin B\sin C=\sin C\sin(A-B)$$
$$\implies \sin B=\sin(A-B)\text{ as }\sin C\ne0$$
$$\implies  B=n\pi+(-1)^n(A-B)\text{  where }n\text{ is any integer} $$
If $n$ is even, $n=2m$(say) $\implies B=2m\pi+A-B\iff A=2B-2m\pi=2B$ as $0<A,B<\pi$
If $n$ is odd, $n=2m+1$(say) $\implies B=(2m+1)\pi-(A-B)\iff A=(2m+1)\pi$ which is impossible as $0<A<\pi$
Conversely, if $A=2B$
$\displaystyle\implies a^2-b^2=4R^2(\sin^2A-\sin^2B)=4R^2\sin C\sin(A-B)$ (using $(1)$)
$\displaystyle=4R^2\sin C\sin(2B-B)=2R\sin B\cdot 2R\sin C=\cdots$
