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
						

							
Dear student 
For short explanation please follow the below Link 
https://www.askiitians.com/forums/Algebra/x-b-c-y-c-a-z-a-bthen-prove-that-b-c-x-c-a-y-a_257383.htm
Hope this help 
Good Luck