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Answer 1

ANSWER:

Given:

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

To Prove:

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

Proof:

$:\longrightarrow\dfrac{y+z-x}{b+c-a}=\dfrac{z+x-y}{c+a-b}=\dfrac{x+y-z}{a+b-c}\\\\\text{Let these fractions be equal to K. That is,}\\\\:\implies\dfrac{y+z-x}{b+c-a}=\dfrac{z+x-y}{c+a-b}=\dfrac{x+y-z}{a+b-c}=K\\\\\text{So,}\\\\:\implies\dfrac{y+z-x}{b+c-a}=K\implies(y+z-x)=K(b+c-a)- - - -(1)\\\\:\implies\dfrac{z+x-y}{c+a-b}=K\implies(z+x-y)=K(c+a-b)- - - -(2)\\\\:\implies\dfrac{x+y-z}{a+b-c}=K\implies(x+y-z)=K(a+b-c)- - - -(3)\\\\\text{Now,}$

$\text{Adding (1) and (2),}\\\\:\implies(y+z-x)+(z+x-y)=K(b+c-a)+K(c+a-b)\\\\:\implies x-x+y-y+z+z=K(a-a+b-b+c+c)\\\\:\implies2z=K(2c)\\\\:\implies z=Kc\\\\:\implies\dfrac{z}{c}=K- - - -(4)\\\\\text{And,}\\\\\text{Adding (1) and (3),}\\\\:\implies(y+z-x)+(x+y-z)=K(b+c-a)+K(a+b-c)\\\\:\implies x-x+y+y+z-z=K(a-a+b+b+c-c)\\\\:\implies2y=K(2b)\\\\:\implies y=Kb\\\\:\implies\dfrac{y}{b}=K- - - -(5)\\\\\text{And,}$

$\text{Adding (2) and (3),}\\\\:\implies(z+x-y)+(x+y-z)=K(c+a-b)+K(a+b-c)\\\\:\implies x+x+y-y+z-z=K(a+a+b-b+c-c)\\\\:\implies2x=K(2a)\\\\:\implies x=Ka\\\\:\implies\dfrac{x}{a}=K- - - -(6)\\\\\text{From (4), (5) and (6),}\\\\:\implies\dfrac{z}{c}=\dfrac{y}{b}=\dfrac{x}{a}\\\\\text{Therefore,}\\\\\bf{:\implies\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}}\\\\\text{HENCE PROVED!!}$

Answer 2

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