In $\triangle ABC$ with biggest angle $A$ and smallest angle $C$, if $A=2C$ and $a+c=2b$, find $a:b:c$

In $$\triangle ABC$$, $$A$$ is the biggest angle, and $$C$$ is the smallest angle. Let $$a$$, $$b$$, $$c$$ be the length of the sides that are opposite from $$\angle A$$, $$\angle B$$, $$\angle C$$, respectively. If $$\angle A=2\angle C$$ and $$a+c=2b$$, find $$a:b:c$$.

I have no idea how to start, but Law of Sines/Law of Cosines could be useful.

$$\frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}$$
This implies that $$\frac{b}{\sin(B)}=\frac{a+c}{\sin(A)+\sin(C)}=\frac{2b}{\sin(A)+\sin(C)}$$
Combine this with $$A=2C\\ B=180^\circ -A -C =180^\circ -3C$$ and you can find $$C$$. From there it is easy
• @Baker013273213 If $\frac{a}{b}=\frac{c}{d}$ then they are also equal to $\frac{a+c}{b+d}$. This is sometimes called "Derived proportions". If you are not familiar, here is the proof: Let $x= \frac{a}{b}=\frac{c}{d}$. Then $$a=bx \\c=dx$$ and hence $$\frac{a+c}{b+d}=\frac{bx+dx}{b+d}=x$$ – N. S. Sep 27 at 2:17
• Continuing on your progress, $\frac{1}{Sin{180-3C}}=\frac{2}{Sin {2C} +Sin {C}$ $\frac{1}{Sin{3C}}=\frac{2}{2 Sin {C} Cos {C} +Sin {C}$ How could we solve this? – Baker013273213 Sep 27 at 19:50