Question

If $\frac{y+z}{pb+qc}=\frac{z+x}{pc+qa}=\frac{x+y}{pa+qb},$ show that $\frac{2(x+y+z)}{a+b+c}=\frac{(b+c)x+(c+a)y+(a+b)z}{bc+ca+ab}.$

Answer 1

$\frac{y+z}{pb+qc}=\frac{z+x}{pc+qa}=\frac{x+y}{pa+qb}=k$

$\therefore (y+z)=k(pb+qc)⟶\left(I\right)$

$(z+x)=k(pc+qa)⟶\left(II\right)$

$(x+y)=k(pa+qb)⟶\left(III\right)$

Adding $1,2$ and $3$

$2(x+y+z)=k[p(a+b+c)+q(a+b+c)]$

$\frac{2(x+y+z)}{a+b+c}=k(p+q)⟶\left(a\right)$

Now, $\left(I\right)\times a$

$\left(II\right)\times b$

$\left(III\right)\times c$ and then adding each equation

$a(y+z)=k(pab+qac)$

$b(z+x)=k(pbc+qab)$

$c(x+y)=k(pac+qbc)$

then add

$⟹(b+c)x+(a+c)y+(a+b)z=k(p+q)[ab+bc+ca]$

$⟹(b+c)x+(a+c)y+(a+b)z=k(p+q)⟶\left(b\right)$

$\because$ Equation $\left(a\right)=\left(b\right)$

$\therefore \frac{2(x+y+z)}{a+b+c}=\frac{(b+c)x+(a+c)y+(a+b)z}{bc+ca+ab}$