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`        If x/b+c=y/c+a=z/a+b then show (b-c) x+(c-a) y+(a-b) z=0`
8 months ago

```							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 - 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) - 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
```
8 months ago
```							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.htmHope this help Good Luck
```
7 months ago
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