Question

(i) If a(y+z) = b(z+x) = c(x+y) and out of a, b, c no two of them are equal then show that,

$\frac{y-z}{a\left(b-c\right)}=\frac{z-x}{b\left(c-a\right)}=\frac{x-y}{c\left(a-b\right)}$


(ii) If $\frac{x}{3x-y-z}=\frac{y}{3y-z-x}=\frac{z}{3z-x-y}$ and$x+y+z\ne 0$ then show that the value of each ratio is equal to 1.


(iii) If $\frac{ax+by}{x+y}=\frac{bx+az}{x+z}=\frac{ay+bz}{y+z}$ and $x+y+z\ne 0$ then show that $\frac{a+b}{2}$.

(iv) If $\frac{y+z}{a}=\frac{z+x}{b}=\frac{x+y}{c}$ then show that $\frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{z}{a+b-c}$ .

(v) If $\frac{3x-5y}{5z+3y}=\frac{x+5z}{y-5x}=\frac{y-z}{x-z}$ then show that every ratio = $\frac{x}{y}$.

Answer 1

(i)
Let $a\left(y+z\right)=b\left(z+x\right)=c\left(x+y\right)=k$

$$\begin{gathered} \Rightarrow a=\frac{k}{y+z},b=\frac{k}{z+x},c=\frac{k}{x+y} \\ \therefore a\left(b-c\right)=\frac{k}{\left(y+z\right)}\left(\frac{k}{z+x}-\frac{k}{x+y}\right) \\ \Rightarrow a\left(b-c\right)=\frac{k^2}{\left(y+z\right)}\left[\frac{x+y-z-x}{\left(z+x\right)\left(x+y\right)}\right] \\ \Rightarrow \frac{y-z}{a\left(b-c\right)}=\frac{k^2}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}.....\left(1\right) \end{gathered}$$
Similarly,

$$\begin{gathered} \frac{z-x}{b\left(c-a\right)}=\frac{k^2}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}.....\left(2\right) \\ \frac{x-y}{c\left(a-b\right)}=\frac{k^2}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}.....\left(3\right) \end{gathered}$$
From (1), (2) and (3), we have

$\frac{y-z}{a\left(b-c\right)}=\frac{z-x}{b\left(c-a\right)}=\frac{x-y}{c\left(a-b\right)}$

(ii)
$\frac{x}{3x-y-z}=\frac{y}{3y-z-x}=\frac{z}{3z-x-y}=\frac{x+y+z}{\left(3x-y-z\right)+\left(3y-z-x\right)+\left(3z-x-y\right)}$ (Theorem of equal ratios)

$\Rightarrow \frac{x}{3x-y-z}=\frac{y}{3y-z-x}=\frac{z}{3z-x-y}=\frac{x+y+z}{x+y+z}$

$\Rightarrow \frac{x}{3x-y-z}=\frac{y}{3y-z-x}=\frac{z}{3z-x-y}=1$

(iii)
$\frac{ax+by}{x+y}=\frac{bx+az}{x+z}=\frac{ay+bz}{y+z}=\frac{\left(ax+by\right)+\left(bx+az\right)+\left(ay+bz\right)}{\left(x+y\right)+\left(x+z\right)+\left(y+z\right)}$ (Theorem of equal ratios)

$$\begin{gathered} \Rightarrow \frac{ax+by}{x+y}=\frac{bx+az}{x+z}=\frac{ay+bz}{y+z}=\frac{\left(a+b\right)x+\left(a+b\right)y+\left(a+b\right)z}{2x+2y+2z} \\ \Rightarrow \frac{ax+by}{x+y}=\frac{bx+az}{x+z}=\frac{ay+bz}{y+z}=\frac{\left(a+b\right)\left(x+y+z\right)}{2\left(x+y+z\right)} \\ \Rightarrow \frac{ax+by}{x+y}=\frac{bx+az}{x+z}=\frac{ay+bz}{y+z}=\frac{a+b}{2} \end{gathered}$$
(iv)
$\frac{y+z}{a}=\frac{z+x}{b}=\frac{x+y}{c}$

$\frac{a}{y+z}=\frac{b}{z+x}=\frac{c}{x+y}$ (By invertendo)

$$\begin{gathered} \frac{a}{y+z}=\frac{b}{z+x}=\frac{c}{x+y}=\frac{b+c-a}{z+x+x+y-y-z}\left(\mathrm{Theorem}\mathrm{of}\mathrm{equal}\mathrm{ratios}\right) \\ \Rightarrow \frac{a}{y+z}=\frac{b}{z+x}=\frac{c}{x+y}=\frac{b+c-a}{2x}.....\left(1\right) \end{gathered}$$
Now,

$$\begin{gathered} \frac{a}{y+z}=\frac{b}{z+x}=\frac{c}{x+y}=\frac{c+a-b}{x+y+y+z-z-x}\left(\mathrm{Theorem}\mathrm{of}\mathrm{equal}\mathrm{ratios}\right) \\ \Rightarrow \frac{a}{y+z}=\frac{b}{z+x}=\frac{c}{x+y}=\frac{c+a-b}{2y}.....\left(2\right) \end{gathered}$$
Also,

$$\begin{gathered} \frac{a}{y+z}=\frac{b}{z+x}=\frac{c}{x+y}=\frac{a+b-c}{y+z+z+x-x-y}\left(\mathrm{Theorem}\mathrm{of}\mathrm{equal}\mathrm{ratios}\right) \\ \Rightarrow \frac{a}{y+z}=\frac{b}{z+x}=\frac{c}{x+y}=\frac{a+b-c}{2z}.....\left(3\right) \end{gathered}$$
From (1), (2) and (3), we have

$$\begin{gathered} \frac{b+c-a}{2x}=\frac{c+a-b}{2y}=\frac{a+b-c}{2z} \\ \Rightarrow \frac{{\displaystyle b+c-a}}{{\displaystyle x}}=\frac{{\displaystyle c+a-b}}{{\displaystyle y}}=\frac{{\displaystyle a+b-c}}{{\displaystyle z}} \\ \Rightarrow \frac{{\displaystyle x}}{{\displaystyle b+c-a}}=\frac{{\displaystyle y}}{{\displaystyle c+a-b}}=\frac{{\displaystyle z}}{{\displaystyle a+b-c}}\left(\mathrm{By}\mathrm{invertendo}\right) \end{gathered}$$
(v)
$$\begin{gathered} \frac{3x-5y}{5z+3y}=\frac{x+5z}{y-5x}=\frac{y-z}{x-z}=\frac{5\left(y-z\right)}{5\left(x-z\right)} \\ \Rightarrow \frac{3x-5y}{5z+3y}=\frac{x+5z}{y-5x}=\frac{5\left(y-z\right)}{5\left(x-z\right)}=\frac{\left(3x-5y\right)+\left(x+5z\right)+5\left(y-z\right)}{\left(5z+3y\right)+\left(y-5x\right)+5\left(x-z\right)}\left(\mathrm{Theorem}\mathrm{of}\mathrm{equal}\mathrm{ratios}\right) \\ \Rightarrow \frac{3x-5y}{5z+3y}=\frac{x+5z}{y-5x}=\frac{y-z}{x-z}=\frac{3x-5y+x+5z+5y-5z}{5z+3y+y-5x+5x-5z} \\ \Rightarrow \frac{3x-5y}{5z+3y}=\frac{x+5z}{y-5x}=\frac{y-z}{x-z}=\frac{4x}{4y} \\ \Rightarrow \frac{3x-5y}{5z+3y}=\frac{x+5z}{y-5x}=\frac{y-z}{x-z}=\frac{x}{y} \end{gathered}$$