Question

if, x/b+c = y/c+a = z/a+b then show that a/y+x-z = b/z+x-y = c/x+y-z​

Answer 1

$\large\textrm{The given equation is proven true.}$

$\Large\textbf{\boxed{\rm{Question}}}$

$\textbf{Prove that:-}$

$\dfrac{x}{y+z-x}=\dfrac{b}{z+x-y}=\dfrac{c}{x+y-z}$

$\textbf{if \dfrac{x}{b+c}=\dfrac{y}{c+a}=\dfrac{z}{a+b} }$

$\Large\boxed{\rm{Concept}}$

$\Large\textbf{Ratios}$

$\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{e}{f}=k$

$a=bk, c=dk, e=fk$

$\cdots\longrightarrow\boxed{\dfrac{a+c-e}{b+d-f}=\dfrac{e+a-c}{f+b-d}=\dfrac{c+e-a}{d+f-b}=k}$

It is an idea, not a formula. In simple words, we can apply the same calculation to both the numerator and denominator without the value changing.

$\Large\boxed{\rm{Solution}}$

$\Large\textbf{Given equation: -}$

$\dfrac{x}{b+c}=\dfrac{y}{c+a}=\dfrac{z}{a+b}$

$\textrm{inverse}$

$\dfrac{b+c}{x}=\dfrac{c+a}{y}=\dfrac{a+b}{z}$

$\small\dfrac{(b+c)+(c+a)-(a+b)}{x+y-z}=\dfrac{(c+a)+(a+b)-(b+c)}{y+z-x}=\dfrac{(a+b)+(b+c)-(c-a)}{z+x-y}$

$\dfrac{2c}{x+y-z}=\dfrac{2a}{y+z-x}=\dfrac{2b}{z+x-y}$

$\dfrac{c}{x+y-z}=\dfrac{a}{y+z-x}=\dfrac{b}{z+x-y}$

$\textrm{rearrange}$

$\dfrac{a}{y+z-x}=\dfrac{b}{z+x-y}=\dfrac{c}{x+y-z}$

$\Large\boxed{\rm{Result}}$

Hence it is proven that the given equation is true.

$\Large\boxed{\text{Learn more}}$

$\Large\textbf{Ratios}$

The concept leads to some important rules.

$\Large\textbf{Componendo}$

$\cdots\longrightarrow\boxed{\textrm{ \dfrac{a}{b}=\dfrac{c}{d} then \dfrac{a+b}{b}=\dfrac{c+d}{d} }}$

$\Large\textbf{Dividendo}$

$\cdots\longrightarrow\boxed{\textrm{ \dfrac{a}{b}=\dfrac{c}{d} then \dfrac{a-b}{b}=\dfrac{c-d}{d} }}$

$\Large\textbf{Componendo - Dividendo}$

$\cdots\longrightarrow\boxed{\textrm{ \dfrac{a}{b}=\dfrac{c}{d} then \dfrac{a+b}{a-b}=\dfrac{c+d}{c-d} }}$

$\cdots\longrightarrow\boxed{\textrm{ \dfrac{a}{b}=\dfrac{c}{d} then \dfrac{a-b}{a+b}=\dfrac{c-d}{c+d} }}$

We can solve questions relating to ratios very effortlessly.

$\textbf{Solve: -}$

$\dfrac{\sqrt{3x+4}+\sqrt{3x-5}}{\sqrt{3x+4}-\sqrt{3x-5}}=9$

$\dfrac{\sqrt{3x+4}+\sqrt{3x-5}}{\sqrt{3x+4}-\sqrt{3x-5}}=\dfrac{9}{1}$

$\textrm{componendo-dividendo}$

$\dfrac{\sqrt{3x+4}+\sqrt{3x-5}+\sqrt{3x+4}-\sqrt{3x-5}}{\sqrt{3x+4}+\sqrt{3x-5}-\sqrt{3x+4}+\sqrt{3x-5}}=\dfrac{9+1}{9-1}$

$\dfrac{2\sqrt{3x+4}}{2\sqrt{3x-5}}=\dfrac{10}{8}$

$\sqrt{\dfrac{3x+4}{3x-5}}=\dfrac{5}{4}$

$\dfrac{3x+4}{3x-5}=\dfrac{25}{16}$

$25(3x-5)=16(3x+4)$

$75x-125=48x+64$

$75x-48x=64+125$

$27x=189$

$x=7$

Some might solve this by rationalization. However, without using rationalization, we solved the question effortlessly. So, this technique is worth knowing.