Question

If $\frac{y}{b+c-a}=\frac{z}{c+a-b}=\frac{x}{a+b-c}$ and
$x+y+z\ne 0$, then which of the following hold true?

• A. $\frac{a}{z+x}=\frac{b}{x+y}=\frac{c}{y+z}$
• B. $\frac{a}{z-x}=\frac{b}{x-y}=\frac{c}{y-z}$
• C. $\frac{z-x}{a}=\frac{x-y}{b}=\frac{y-z}{c}$
• D. $\frac{z+x}{a}=\frac{b}{x+y}=\frac{c}{y+z}$

Answer 1

The correct option is A $\frac{a}{z+x}=\frac{b}{x+y}=\frac{c}{y+z}$

Given:
$\frac{y}{b+c-a}=\frac{z}{c+a-b}=\frac{x}{a+b-c}$

By invertendo, we get

$\frac{b+c-a}{y}=\frac{c+a-b}{z}=\frac{a+b-c}{x}$
Now, let $\frac{b+c-a}{y}=\frac{c+a-b}{z}=\frac{a+b-c}{x}=k$
$\therefore$ by theorem of equal ratio:
$\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}$

$\Rightarrow k=\frac{(c+a-b)+(a+b-c)}{z+x}$
$\Rightarrow k=\frac{2a}{z+x}-----\left(I\right)$
Similarly, it can be written for other terms:

$k=\frac{(a+b-c)+(b+c-a)}{x+y}$
$\Rightarrow k=\frac{2b}{x+y}-----\left(II\right)$

$k=\frac{(b+c-a)+(c+a-b)}{y+z}$
$\Rightarrow k=\frac{2c}{y+z}-----\left(II\right)$
Using equation $I,II\text{and}III$,
$\therefore \frac{2a}{z+x}=\frac{2b}{x+y}=\frac{2c}{y+z}$
$\therefore \frac{a}{z+x}=\frac{b}{x+y}=\frac{c}{y+z}$