Question

1 / b+c + 1/c+a = 2/ a+b, then the value of a^2 + b^2 / c2 is

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: \dfrac{1}{b + c} + \dfrac{1}{c + a} = \dfrac{2}{a + b}$

can be rewritten as

$\rm \: \dfrac{1}{b + c} + \dfrac{1}{c + a} = \dfrac{1}{a + b} + \dfrac{1}{a + b}$

can be further rearrange as

$\rm \: \dfrac{1}{a + b} - \dfrac{1}{b + c} = \dfrac{1}{c + a} - \dfrac{1}{a + b}$

$\rm \: \dfrac{b + c - (a + b)}{(a + b)(b + c)} = \dfrac{a + b - (c + a)}{(c + a)(a + b)}$

$\rm \: \dfrac{b + c - a - b}{(b + c)} = \dfrac{a + b - c - a}{(c + a)}$

$\rm \: \dfrac{c - a}{b + c} = \dfrac{b - c}{c + a}$

$\rm \: (c + a)(c - a) = (b + c)(b - c)$

$\rm \: {c}^{2} - {a}^{2} = {b}^{2} - {c}^{2}$

$\rm \:\bigg [ \because \: (x + y)(x - y) = {x}^{2} - {y}^{2} \: \bigg]$

$\rm \: {c}^{2} + {c}^{2} = {b}^{2} + {a}^{2}$

$\rm \: 2{c}^{2} = {a}^{2} + {b}^{2}$

$\bf\implies \:\boxed{ \bf{ \:\dfrac{ {a}^{2} + {b}^{2} }{ {c}^{2} } = 2 \: }}$