<s>[INST] 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.
 [/INST] Double angle formula says
$$
\begin{align}
\sin(A)
&=2\sin(B)\cos(B)\\
&\implies\frac{\sin(A)}{\sin(B)}=2\cos(B)\tag{1}
\end{align}
$$
The formula for the sine of a sum yields
$$
\begin{align}
\sin(C)
&=\sin(A+B)\\
&=2\sin(B)\cos(B)\cos(B)+(2\cos^2(B)-1)\sin(B)\\
&=\sin(B)(4\cos^2(B)-1)\\
&\implies\frac{\sin(C)}{\sin(B)}=4\cos^2(B)-1\tag{2}
\end{align}
$$
Thus, the Law of Sines says
$$
\left(\frac ab\right)^2-1=\frac cb\implies a^2-b^2=bc\tag{3}
$$
</s>