Question

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

Answer 1

Let $\frac{x}{b+c}=\frac{y}{c+a}=\frac{z}{a+b}=K$

$\therefore x=K(b+c),y=K(c+a),Z=K(a+b)$

$LHS=(b-c)x+(c-a)y+(a-b)z$

Putting values of $x,y$ and $z$.

$=(b-c)K(b+c)+(c-a)K(c+a)+(a-b)K(a+b)$

$=K\left[\right(b-c\left)\right(b+c)+(c-a\left)\right(c+a)+(a-b\left)\right(a+b\left)\right]$

$=K[b^2+bc-bc-c^2+c^2+ca-ca-a^2+a^2+ab-ab-b^2]$

$=K[b^2-c^2+c^2-a^2+a^2-b^2+bc-bc+ca-ca+ab-ab]$

$=K\left[0\right]$

$=0$

$=RHS$

Hence, proved.