Question

Solve this....
I will mark u brainlist​

Answer 1

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

Given that

$\rm :\longmapsto\:\dfrac{by + cz}{ {b}^{2} + {c}^{2} } = \dfrac{cz + ax}{ {c}^{2} + {a}^{2} } = \dfrac{ax + by}{ {a}^{2} + {b}^{2} }$

We know, if

$\boxed{ \bf{ \: \dfrac{a}{b} = \dfrac{c}{d} \rm \implies\:\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{a + b}{c + d} }}$

This process is called Addendo.

So, using this process, we get

$\rm :\longmapsto\:\dfrac{by + cz}{ {b}^{2} + {c}^{2} } = \dfrac{cz + ax}{ {c}^{2} + {a}^{2} } = \dfrac{ax + by}{ {a}^{2} + {b}^{2} } = \dfrac{2ax + 2by + 2cz}{2 {a}^{2}+{2b}^{2}+{2c}^{2} }$

$\rm :\longmapsto\:\dfrac{by + cz}{ {b}^{2} + {c}^{2} } = \dfrac{cz + ax}{ {c}^{2} + {a}^{2} } = \dfrac{ax + by}{ {a}^{2} + {b}^{2} } = \dfrac{2(ax + by + cz)}{2 ({a}^{2}+{b}^{2}+{c}^{2}) }$

$\rm :\longmapsto\:\dfrac{by + cz}{ {b}^{2} + {c}^{2} } = \dfrac{cz + ax}{ {c}^{2} + {a}^{2} } = \dfrac{ax + by}{ {a}^{2} + {b}^{2} } = \dfrac{ax + by + cz}{{a}^{2}+{b}^{2}+{c}^{2}}$

Now, Consider first and fourth member

$\rm :\longmapsto\:\dfrac{by + cz}{ {b}^{2} + {c}^{2} } = \dfrac{ax + by + cz}{{a}^{2}+{b}^{2}+{c}^{2}}$

can be rewritten as

$\rm :\longmapsto\:\dfrac{ {a}^{2} + {b}^{2} + {c}^{2} }{ {b}^{2} + {c}^{2} } = \dfrac{ax + by + cz}{by + cz}$

On Subtracting 1 from each term, we get

$\rm :\longmapsto\:\dfrac{ {a}^{2} + {b}^{2} + {c}^{2} }{ {b}^{2} + {c}^{2} } - 1 = \dfrac{ax + by + cz}{by + cz} - 1$

$\rm :\longmapsto\:\dfrac{ {a}^{2} + {b}^{2} + {c}^{2} - {b}^{2} - {c}^{2} }{ {b}^{2} + {c}^{2} } = \dfrac{ax + by + cz - by - cz}{by + cz}$

$\rm :\longmapsto\:\dfrac{ {a}^{2}}{ {b}^{2} + {c}^{2} } = \dfrac{ax }{by + cz}$

On cancelation of a in numerators, we get

$\rm :\longmapsto\:\dfrac{ {a}}{ {b}^{2} + {c}^{2} } = \dfrac{x }{by + cz}$

$\rm :\longmapsto\:\dfrac{ {x}}{a} = \dfrac{by + cz}{ {b}^{2} + {c}^{2} } - - - (1)$

Now, Consider second and fourth member, we get

$\rm :\longmapsto\: \dfrac{cz + ax}{ {c}^{2} + {a}^{2} } = \dfrac{ax + by + cz}{{a}^{2}+{b}^{2}+{c}^{2}}$

can be rewritten as

$\rm :\longmapsto\:\dfrac{ {a}^{2} + {b}^{2} + {c}^{2} }{ {c}^{2} + {a}^{2} } = \dfrac{ax + by + cz}{cz + ax}$

On Subtracting 1 from each term, we get

$\rm :\longmapsto\:\dfrac{ {a}^{2} + {b}^{2} + {c}^{2} }{ {c}^{2} + {a}^{2} } - 1 = \dfrac{ax + by + cz}{cz + ax} - 1$

$\rm :\longmapsto\:\dfrac{ {a}^{2} + {b}^{2} + {c}^{2} - {c}^{2} - {a}^{2} }{ {c}^{2} + {a}^{2} } = \dfrac{ax + by + cz - cz - ax}{cz + ax}$

$\rm :\longmapsto\:\dfrac{ {b}^{2}}{ {c}^{2} + {a}^{2} } = \dfrac{by }{cz + ax}$

On cancel b in numerators, we get

$\rm :\longmapsto\:\dfrac{ {b}}{ {c}^{2} + {a}^{2} } = \dfrac{y }{cz + ax}$

can be rewritten as

$\rm :\longmapsto\:\dfrac{ {cz + ax}}{ {c}^{2} + {a}^{2} } = \dfrac{y }{b} - - - - (2)$

Now, Consider third and fourth member,

$\rm :\longmapsto\: \dfrac{ax + by}{ {a}^{2} + {b}^{2} } = \dfrac{ax + by + cz}{{a}^{2}+{b}^{2}+{c}^{2}}$

can be rewritten as

$\rm :\longmapsto\:\dfrac{ {a}^{2} + {b}^{2} + {c}^{2} }{ {b}^{2} + {a}^{2} } = \dfrac{ax + by + cz}{by + ax}$

On Subtracting 1 from each term, we get

$\rm :\longmapsto\:\dfrac{ {a}^{2} + {b}^{2} + {c}^{2} }{ {b}^{2} + {a}^{2} } - 1 = \dfrac{ax + by + cz}{by + ax} - 1$

$\rm :\longmapsto\:\dfrac{ {a}^{2} + {b}^{2} + {c}^{2} - {b}^{2} - {a}^{2} }{ {b}^{2} + {a}^{2} }= \dfrac{ax + by + cz - by - ax}{by + ax}$

$\rm :\longmapsto\:\dfrac{ {c}^{2} }{ {b}^{2} + {a}^{2} }= \dfrac{cz}{by + ax}$

$\rm :\longmapsto\:\dfrac{ {c}}{ {b}^{2} + {a}^{2} }= \dfrac{z}{by + ax}$

$\rm :\longmapsto\:\dfrac{ {ax + by}}{ {a}^{2} + {b}^{2} }= \dfrac{z}{c} - - - - (3)$

As it is given that,

$\rm :\longmapsto\:\dfrac{by + cz}{ {b}^{2} + {c}^{2} } = \dfrac{cz + ax}{ {c}^{2} + {a}^{2} } = \dfrac{ax + by}{ {a}^{2} + {b}^{2} }$

From equation (1), (2) and (3), we concluded that

$\bf\implies \:\dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c}$

Hence, Proved