Question

x/b+c = y/c+a = z/a+b . so prove that (b-c)x + (c-a)y + (a-b) z = 0

Answer 1

I think you have done a mistake in the given condition.

x/(b+c-a) = y/( c+a-b) = z/(a+b-c) = k

There are three fractions.

Use the following properties of fractions.

1. We know that if both numerator and denominator of a fraction p/q are multiplied by the same term, it remains the same

p/q = px / qx = py / qy = pz/qz

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2. a/b = c/d = e/f

Then adding all the numerators and denominators, the fraction doesn't change.

a/b = c/d = e/f = (a + c + e) / (b + d + f)

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Using above two properties together

Multiply first fraction by (b - c)/(b - c), second by (c -a)/(c - a), third by (a - b)/(a - b) then adding them

Numerator = (b - c)x + (c -a)y + (a - b)z

Denominator = (b - c)(b + c - a) + (c - a)(c + a - b) + (a - b)(a + b - c)

Denominator = b2 - c2 - a(b - c) + c2 - a2- b(c - a) + a2 - b2 - c(a - b)

Denominator = - a(b - c)- b(c - a) - c(a - b)

Denominator = - ab + ac - bc + ab - ca +bc

Denominator = 0

Therefore

x/(b+c-a) = y/( c+a-b) = z/(a+b-c) = [(b - c)x + (c -a)y + (a - b)z ] / 0 = k

(b - c)x + (c -a)y + (a - b)z = 0

Answer is zero