Question

If a(y+z-x)= b(z+x-y)= c(x+y-z), then show that x(b+c)= y(c+a)= z(a+b)​

Answer 1

Step-by-step explanation:

given,  a(y+z-x) = b(z+x-y) = c(x+y-z)  = k (assume)

=> a(y+z-x) = k    => a = k/(y+z-x)

=> b(z+x-y) = k     => b = k/(z+x-y)

=> c(x+y-z)  = k    => c = k/(x+y-z)

consider (a+b)

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

=> a+b = k [  $\frac{1}{(y+z-x)}$ + $\frac{1}{(z+x-y)}$ ]  

=> a+b = k [ $\frac{(z+x-y)+(y+z-x)}{(y+z-x)(z+x-y)}$ ]

=> a+b = k [ $\frac{(z+x-y)+(y+z-x)}{(y+z-x)(z+x-y)}$ ]

=> a+b = k [ $\frac{2z}{(z- (x-y))(z+(x-y))}$ ]

=> a+b = k [ $\frac{2z}{z^2- (x-y)^2}$ ] = k [ $\frac{2z}{z^2- x^2-y^2+2xy}$ ]

=> (a+b) [ $\frac{z^2- x^2-y^2+2xy}{2z}$ ] = k  ---------------- (1)

consider (b+c)

=> b+c =  k/(z+x-y) + k/(x+y-z)

=> b+c = k [  $\frac{1}{(z+x-y)}$ +$\frac{1}{(x+y-z)}$  ]  

=> b+c = k [ $\frac{(z+x-y)+(x+y-z)}{(x+y-z)(z+x-y)}$ ]

=> b+c = k [ $\frac{2x}{(x+(y-z))(x-(y-z))}$ ]

=> b+c = k [ $\frac{2x}{x^2- (y-z)^2}$ ] = k [ $\frac{2x}{x^2- y^2-z^2+2yz}$ ]

=> (b+c) [ $\frac{x^2- y^2-z^2+2yz}{2x}$ ] = k  ---------------- (2)

consider (a+c)

=> a+c =  k/(y+z-x) + k/(x+y-z)

=> a+c = k [  $\frac{1}{(y+z-x)}$ +$\frac{1}{(x+y-z)}$  ]  

=> a+c = k [ $\frac{(y+z-x)+(x+y-z)}{(x+y-z)(y+z-x)}$ ]

=> a+c = k [ $\frac{2y}{(y+(x-z))(y-(x-z))}$ ]

=> a+c = k [ $\frac{2y}{y^2- (x-z)^2}$ ]  = k [ $\frac{2y}{y^2- x^2-z^2+2xz}$ ]

=> (a+c) [ $\frac{y^2- x^2-z^2+2xz}{2y}$ ] = k ----------------------(3)

by (1) and (2) =>

(a+b) [ $\frac{z^2- x^2-y^2+2xy}{2z}$ ] = (b+c) [ $\frac{x^2- y^2-z^2+2yz}{2x}$ ]

=> $\frac{a+b}{b+c}$  = $\frac{(x+y-z) (x-y+z) )}{2x}$ * $\frac{2z}{(y+z- x))(x+z-y)}$

=> $\frac{a+b}{b+c}$  = $\frac{(x+y-z) (x-y+z) )}{2x}$ * $\frac{2z}{(y+z- x))(x+z-y)}$ = $\frac{z(x+y-z)}{x(y+z- x)}$

=> $\frac{a+b}{b+c}$ = $\frac{x}{z}$

=> z(a+b) = x(b+c)

similarly, we can prove,

z(a+b)  = y(a+c)

=> x(b+c) = y(a+c) = z(a+b)