Question

ALZEBRA IMPORTANT QUESTION CLASS 9

Answer 1

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


From the properties of ratio and proportion, we know : -

$\mathsf{Each \: Ratio =\dfrac{Sum\: of \: antecedents}{Sum\: of \: consequents}}$


Therefore,

$Each \: ratio = \dfrac{(a + b - c ) + (b + c - a) + (a + c - b)}{a + b + b + c + a+ c} \\ \\ \\ \implies Each \: ratio = \dfrac{a + b - c + b + c - a+ a + c - b}{a + b + b + c + a+ c} \\ \\ \\ \implies \: Each \: ratio = \dfrac{a + b + c}{2(a + b + c)} \\ \\ \\ \implies Each \: ratio = \dfrac{1}{2}$



Now,
$\dfrac{a + b - c}{a + b} = \dfrac{b + c - a}{b + c} = \dfrac{c + a - b}{c + a} = \dfrac{1}{2}$



Therefore,
• ( a + b - c ) / ( a + b ) = 1 / 2

= > 2a + 2b - 2c = a + b

= > 2a - a + 2b - b = 2c

= > a + b = 2c

= > a = 2c - b ---: ( 1 )



• ( b + c - a ) / ( b + c ) = 1 / 2

= > 2b + 2c - 2a = b + c

= > 2b - b + 2c - c = 2a

= > b + c = 2a


Substituting the value of a from ( 1 ),

= > b + c = 2( 2c - b )

= > b + c = 4c - 2b

= > b + 2b = 4c - c

= > 3b = 3c

= > b = c ---: ( 2 )


Now, substituting the value of b in ( 1 ),

= > a = 2c - b

= > a = 2c - c

= > a = c ---: ( 3 )




Then, comparing ( 1 ) & ( 2 ),

= > b = c = a


Proved.
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