Prove a rational fraction is equal to another

[brotherbobby]

Homework Statement

If $\dfrac{x}{b+c-a}=\dfrac{y}{c+a-b}=\dfrac{z}{a+b-c},$ $\\[10pt]$
prove that $\boxed{\boldsymbol{\dfrac{x+y+z}{a+b+c}=\dfrac{x(y+z)+y(z+x)+z(x+y)}{2(ax+by+cz)}}}$

Relevant Equations

$\mathbf{Theorem :}$ If fractions $\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{e}{f}=\ldots$, each of the fractions is equal to $\left( \dfrac{pa^n+qc^n+re^n+\ldots}{pb^n+qd^n+rf^n+\ldots}\right)^{1/n}$

Problem Statement : If $\dfrac{x}{b+c-a}=\dfrac{y}{c+a-b}=\dfrac{z}{a+b-c},$ prove that $\boxed{\boldsymbol{\dfrac{x+y+z}{a+b+c}=\dfrac{x(y+z)+y(z+x)+z(x+y)}{2(ax+by+cz)}}}$

Attempt : Let the fractions (ratios) $\dfrac{x}{b+c-a}=\dfrac{y}{c+a-b}=\dfrac{z}{a+b-c} = \boldsymbol{k}$. $\\[10pt]$
One consequnce of the Relevant equations given above is that if $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=\ldots = \frac{a+c+e+\ldots}{b+d+f+\ldots}$. This can be shown by putting $p=q=r=\ldots = n = 1$. $\\[10 pt]$
Hence from the problem statement $\dfrac{x}{b+c-a}=\dfrac{y}{c+a-b}=\dfrac{z}{a+b-c} = \dfrac{x+y+z}{b+c-a+c+a-b+a+b-c}= \dfrac{x+y+z}{a+b+c}$.

So the L.H.S of the required equation is the same as each of the given fractions.

Hence, what we need to show is that the R.H.S of the required equation is also equal to the given fractions, which I have simply put to be equal to $k$.

Let's see. The R.H.S. = $\dfrac{x(y+z)+y(z+x)+z(x+y)}{2(ax+by+cz)}= \dfrac{(x+y+z)^2-(x^2+y^2+z^2)}{2(ax+by+cz)}=\dfrac{k^2(a+b+c)^2-k^2\{(b+c-a)^2+(c+a-b)^2+(a+b-c)^2\}}{2k\{ a(b+c-a)+b(c+a-b)+c(a+b-c) \}}$ $\\[10pt]$ $=\dfrac{k}{2}\dfrac{-2(a^2+b^2+c^2)+(4bc+4ca+4ab)}{2bc+2ca+2ab-(a^2+b^2+c^2)}=k\dfrac{(a^2+b^2+c^2)-2(ab+bc+ca)}{(a^2+b^2+c^2)-(2ab+2bc+2ca)}= k$ (proved).

The issue : The book I am working from (Hall and Knight, Higher Algebra), has explicitly asked not to assume the given ratios (fractions) $\dfrac{x}{b+c-a}=\dfrac{y}{c+a-b}=\dfrac{z}{a+b-c} = \boldsymbol{k}$, which is what I have done. I can't see right now how the problem above can be done without it.

A hint or a suggestion to solve the problem without the use of the "fractional representative" $k$ would be welcome.

 

[anuttarasammyak]

Using
$$x=(b+c-a)k$$$$y=(c+a-b)k$$$$z=(a+b-c)k$$
and substituting x,y,z with a,b,c,k formula seems a straight forward way. What sort of anxiety you feel on it ?

 

[Delta2]

Maybe it can be done by using the property $$\frac{A}{B}=\frac{C}{D}\iff AD=BC$$ but it is going to be long and painful as you going to use this property many times for the various fractions. Seems to me that the book authors want to do this in the longest, most painful way, but you going to practice your algebra skills doing it this way.

 

[brotherbobby]

Using
$$x=(b+c-a)k$$$$y=(c+a-b)k$$$$z=(a+b-c)k$$
and substituting x,y,z with a,b,c,k formula seems a straight forward way. What sort of anxiety you feel on it ?

That is what I did. But the authors of the book want the problem solved without the use of the "fractional representative" k.

 

[anuttarasammyak]

Hum... Substitutions without explicit k,
$$x=\frac{b+c-a}{a+b-c}z$$
$$y=\frac{c+a-b}{a+b-c}z$$
do also work but less beautiful. I do not think it is the intention of the author.

 

[brotherbobby]

The authours have uploaded their solution to the problem above. The solution is elegant but requires plenty of practice for one to be aware of it. I am afraid I didn't have it at the time of writing.

Problem statement : If $\dfrac{x}{b+c-a}=\dfrac{y}{c+a-b}=\dfrac{z}{a+b-c}$ $\\[20pt]$, prove that $\boxed{\boldsymbol{\dfrac{x+y+z}{a+b+c}=\dfrac{x(y+z)+y(z+x)+z(x+y)}{2(ax+by+cz)}}}$.

Solution (authors) :