Triangle proof using law of sines 
In triangle $ABC$, suppose that angle $C$ is twice angle $A$. Use the law of sines to show that $ab= c^2 - a^2$.

 A: Suggestions for getting started: 


*

*Set $\angle C = 2\angle A$. And the measure of angle $B$ would then be $\pi - 3A$. Why?

*Use the law of Sines (or the reciprocal): 
$$ \frac{\sin A}{a} \,=\, \frac{\sin B}{b} \,=\, \frac{\sin C}{c}\quad\;\text{or}\quad
    \frac{a}{\sin A} \,=\, \frac{b}{\sin B} \,=\, \frac{c}{\sin C}$$ Substituting $2A$ for $C$, and use the double angle formula for $\sin C = \sin(2A)$. Substitute, as well $\sin B = \sin(\pi - 3A) = \sin(3A)$ 

*Make all possible substitutions, use trig identities, and simplify, and see where that gets you, with your goal being to relate the lengths of the sides as stated. 
We'll be happy to check your progress!
A: Put $\,\angle C=2w\;,\;\;\angle A=w\;,\;\;\angle B=180- 3w\;$ , then
$$\frac{b}{\sin 3w}=\frac{c}{\sin 2w}=\frac{a}{\sin w}\implies ab=\frac{b^2\sin w}{\sin 3w}{}\;,\;\;a^2-c^2=\frac{b^2\sin^22w}{\sin^33w}-\frac{b^2\sin^2w}{\sin^23w}\implies$$
$$ab=c^2-a^2\iff \frac{b^2}{\sin 3w}\sin w=\frac{b^2}{\sin 3w}\left(\frac{\sin^22w}{\sin 3w}-\frac{\sin^2w}{\sin 3w}\right)\iff$$
$$\iff \sin w=\frac{\sin^22w-\sin^2w}{\sin 3w}=\frac{4\sin^2w\cos^2w-\sin^2w}{\sin 3w}\iff$$
$$\iff \frac{4\sin w\cos^2 w-\sin w}{\sin 3w}=1$$
But
$$(1)\;\;\;4\sin w\cos^2w-\sin w=\sin w(4\cos^2w-1)$$
$$(2)\;\;\;\sin 3w=\sin w\cos 2w+\sin 2w\cos w=\sin w(2\cos^2w-1)+2\sin w\cos^2w$$
Now just check that $\,(1)=(2)\;$...
A: A common way to invoke the Law of Sines is to note what its ratios equal:
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = d$$
where $d$ is the circum-diameter of $\triangle ABC$. Thus, 
$$a = d \sin A \qquad b = d \sin B \qquad c = d \sin C$$
Consequently, the identity in question, upon division by $d^2$, is equivalent to
$$\sin A \sin B \stackrel{?}{=} \sin^2 C - \sin^2 A$$
Now, $C=2A$ implies $B=\pi-3A$, so that $\sin B = \sin 3A$, and the identity to be proven becomes
$$\begin{align}
\sin A \sin 3A &\stackrel{?}{=} \sin^2 2A - \sin^2 A \\[6pt]
&= \left( \sin 2A + \sin A \right) \left( \sin 2 A - \sin A \right) \\[6pt]
&= 2 \sin\frac{3A}{2}\cos\frac{A}{2} \cdot 2 \sin\frac{A}{2} \cos\frac{3A}{2} \\[6pt]
&= 2 \sin\frac{A}{2}\cos\frac{A}{2} \cdot 2 \sin\frac{3A}{2} \cos\frac{3A}{2} \\[6pt]
&= \sin A \; \sin 3A
\end{align}$$
In the above, I use the sum-to-product identity $\sin\theta \pm \sin\phi = 2 \sin\frac{\theta\pm \phi}{2} \cos\frac{\theta\mp \phi}{2}$, which is familiar to me. Someone familiar with multiple-angle identities may have proceeded thusly:
$$\begin{align}
\sin A \sin 3A &\stackrel{?}{=} \sin^2 2A - \sin^2 A \\[6pt]
&= 4\sin^2 A\cos^2 A - \sin^2 A \\[6pt]
&= \sin A \cdot \sin A \left( 4 \cos^2 A - 1 \right) \\[6pt]
&= \sin A \cdot \sin A \left( 3 - 4\sin^2 A \right) \\[6pt]
&= \sin A \; \sin 3A
\end{align}$$
A: Not sure whether I can submit this answer, but now that you know the proof through trigonometry, here's a different approach through simple geometry:

Look at the $\triangle ABC$ and $\triangle CDB$, they are similar(Why?)
Lets call $AB=c$, $AC=b$ and $CB=a$. (As the usual notations)
Now from similarity we have, $\dfrac{b}{CD}=\dfrac{c}{a}=\dfrac{a}{DB}$
You will get the result with some manipulations(Get $DB$ in terms of sides of triangle).
