Geometry Proving Problem

$A=2B, C=\pi-3B$ $\small{\implies \sin A=\sin 2B \text{ and } \sin C=\sin 3B=3\sin B-4\sin^3 B}$

Using $a=2R\sin A, b=2R\sin B ...$ so that statement is: $\sin^2A=\sin B(\sin B+\color{#D61F06}{\sin C})$ $\text{OR } \sin^2 2B=\sin B(\sin B+\color{#D61F06}{3\sin B-4\sin^3 B})$ $LHS: \sin^2B(4-4\sin^2 B)=4\sin^2B\cos^2B\\~~~~~~~~~~~~~=\sin^2 2B=RHS$

Hence statement is $\boxed{\text{True}}$ .

Note: The question is equivalent to "Prove that $BC^2=AC(AC+AB)$ ".

We start by introducing a straight line $AD$ with point $D$ on line $BC$ such that $\angle{DAE}=\angle{DBA}=\theta$ . Now, observe that $\angle{CAD}=\theta$ , $\angle{ADC}=2\theta$ , $\angle{ACD}=180-3\theta$ . Therefore, we can also deduce that $\triangle{ACD}\sim\triangle{ECA}(AAA)$

Then, we can use this similarity to derive the following: $\frac{AC}{BC}=\frac{CD}{AC}=\frac{AD}{AB}\Rightarrow\frac{AC}{AD+CD}=\frac{CD}{AC}=\frac{AD}{AB}\Rightarrow AD\times CD+CD^2=AC^2\text{ and }AD\times CD+AD^2=AB\times AC$

Then, substituting in, we get $BC^2=(AD+CD)^2=AD^2+CD^2+2AD\times CD=AC^2+AC\times AB=AC(AC+AB)$