Question

Anyone who can help please fast​

Answer 1

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

$\rm :\longmapsto\:a : b = c : d$

$\large\underline{\sf{To\:prove - }}$

$\rm :\longmapsto\:\dfrac{ {a}^{2} + ac + {c}^{2} }{ {a}^{2} - ac + {c}^{2} } = \dfrac{ {b}^{2} + bd + {d}^{2} }{ {b}^{2} - bd+ {d}^{2} }$

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

Given that

$\rm :\longmapsto\:a : b = c : d$

It means,

$\rm :\longmapsto\:\dfrac{a}{b} = \dfrac{c}{d}$

Let assume that

$\rm :\longmapsto\:\dfrac{a}{b} = \dfrac{c}{d} = k$

So, it implies

$\rm :\longmapsto\:a = bk$

$\rm :\longmapsto\:c = dk$

Consider,

LHS

$\rm :\longmapsto\:\dfrac{ {a}^{2} + ac + {c}^{2} }{ {a}^{2} - ac + {c}^{2} }$

On substituting the values of a and c, we get

$\rm \: = \: \: \dfrac{ {(bk)}^{2} + (bk)(dk) + {(dk)}^{2} }{{(bk)}^{2} - (bk)(dk) + {(dk)}^{2}}$

$\rm \: = \: \: \dfrac{ {b}^{2} {k}^{2} + bd {k}^{2} + {d}^{2} {k}^{2} }{{b}^{2} {k}^{2} - bd {k}^{2} + {d}^{2} {k}^{2}}$

$\rm \: = \: \: \dfrac{ {k}^{2} ({b}^{2} + bd+ {d}^{2})}{{k}^{2}({b}^{2} - bd + {d}^{2})}$

$\rm \: = \: \: \dfrac{{b}^{2} + bd+ {d}^{2}}{{b}^{2} - bd + {d}^{2}}$

$\boxed{\boxed{\bf{Hence, Proved}}}$

Additional Information :-

$\rm :\longmapsto\:If \: \dfrac{a}{b} = \dfrac{c}{d} , \: then$

$\rm :\longmapsto\:\dfrac{a}{c} = \dfrac{b}{d} \: is \: called \: alternendo$

$\rm :\longmapsto\:\dfrac{b}{a} = \dfrac{d}{c} \: is \: called \: invertendo$

$\rm :\longmapsto\:\dfrac{a + b}{b} = \dfrac{c + d}{d} \: is \: called \: componendo$

$\rm :\longmapsto\:\dfrac{a - b}{b} = \dfrac{c - d}{d} \: is \: called \: dividendo$

$\rm :\longmapsto\:\dfrac{a + b}{a - b} = \dfrac{c + d}{c - d} \: is \: called \:componendo \: and \: dividendo$