Question

If $\frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{z}{a+b-c}.$ then show that x(b $-$ c) + y (c $-$ a) + z (a $-$ b) = 0.

Answer 1

$$\begin{gathered} \mathrm{Let}: \\ \frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{z}{a+b-c}=k \\ \mathrm{Thus},\mathrm{we}\mathrm{have}: \\ x=\left(b+c-a\right)k,y=\left(c+a-b\right)kandz=\left(a+b-c\right)k \\ \mathrm{Now}, \\ \mathrm{LHS}=x\left(b-c\right)+y\left(c-a\right)+z\left(a-b\right) \\ =\left(b+c-a\right)\left(b-c\right)k+\left(c+a-b\right)\left(c-a\right)k+\left(a+b-c\right)\left(a-b\right)k \\ =k\left(b^2+bc-ba-bc-c^2+ac+c^2-ca+ac-a^2-bc+ba+a^2-ab+ba-b^2-ca+cb\right) \\ =k\left(0\right) \\ =0 \\ =\mathrm{RHS} \\ \therefore x\left(b-c\right)+y\left(c-a\right)+z\left(a-b\right)=0 \end{gathered}$$