1. ## Ratio

Hello all,

I need some help on this problem:

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

2. Hi!
You can easily get it by using the property
if a/b = c/d
then a/b = c/d = (a+c)/(b+d)

Try it. If you can't get it, tell me.

Keep Smiling
Malay

3. I tried ...

The sum of $\frac{y+z}{pb+qc} = \frac{z+x}{pc+qa} =\frac{x+y}{pa+qb} = \frac{2(x+y+z)}{(p+q)(a+b+c)}$

It is close to the proof except for $(p+b)$ in the denominator. What should I do next? It can disappear if I assume $(p+b) = 1$

4. $\frac{y+z}{pq+bc}=\frac{ay+az}{pab+qac}$
$\frac{z+x}{pc+qa}=\frac{bz+bx}{pbc+qab}$
$\frac{x+y}{pa+qb}=\frac{cx+cy}{pac+qbc}$
$\frac{y+z}{pq+bc}+\frac{z+x}{pc+qa}+\frac{x+y}{pa+ qb}=\frac{ay+az}{pab+qac}+\frac{bz+bx}{pbc+qab}+\f rac{cx+cy}{pac+qbc}$