Question

If $\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}$, then prove that each ratio is equal to $\frac{1}{2}$ or $-1$.

Answer 1

Given $\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}$

If $\frac{a}{b+c}=\frac{b}{c+a}$

$\Rightarrow ac+a^2=b^2+bc$

$a^2-b^2=bc-ac$

$\Rightarrow (a+b)(a-b)=c(b-a)$

$\Rightarrow a+b=-c$

Then, $\frac{c}{a+b}=\frac{c}{-c}=-1$

If $\frac{b}{c+a}=\frac{c}{a+b}$

$\Rightarrow ab+b^2=c^2+ac$

$\Rightarrow b^2-c^2=ac-ab$

$\Rightarrow (b-c)(b+c)=-a(b-c)$

$\Rightarrow b+c=-a$

Then, $\frac{a}{b+c}=\frac{a}{-a}=-1$

Similarly, $\frac{b}{c+a}=\frac{b}{-b}=-1$

$\therefore \frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=-1$