Difficulty in proving a result

Question

If $\frac{ay-bx}{c}$ = $\frac{cx-az}{b}$ = $\frac{bz-cy}{a}$ , prove $\frac{x}{a}$= $\frac{y}{b}$ = $\frac{z}{c}$.

My attempt:

$\frac{(ay-bx)z}{cz}$ = $\frac{(cx-az)y}{by}$ = $\frac{(bz-cy)x}{ax}$

⇒$\frac{ayz-bxz}{cz}$ = $\frac{cxy-ayz}{by}$ = $\frac{bxz-cxy}{ax}$

Applying componendo and dividendo we get,

$\frac{ayz-bxz + cxy-ayz + bxz-cxy}{cz + by + ax}$ = each of the ratios

⇒0 = each of the ratios

Therefore,

$\frac{ay-bx}{c}$ = 0

⇒$ay - bx = 0$

⇒$ay = bx$

⇒$x/a = y/b$

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$\frac{cx-az}{b} = 0$

⇒$cx-az = 0$

⇒$cx = az$

⇒$z/c = x/a$

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Therefore we get, $\frac{x}{a}$ = $\frac{y}{b}$ = $\frac{z}{c}$

My question: I do not like this method. Is there any better method to prove this result (Preferably which does not uses componendo and dividendo and uses the k-method which is some what more logical). I was trying to prove this by using the k - method but could not proceed far.

My attempt (with the k - method):

Since we have to prove that $\frac{x}{a}$= $\frac{y}{b}$ = $\frac{z}{c}$, our task would be accomplished if we could some how get this equation: $(x-ak)(y-bk) (z-ck) = 0$ (where k is some constant of proportionality, not necessarily equal to the given equation.

After this i multiplied the factors but could not arrive there by using the given equation.

I want a method in which the result can be obtained by comparing the terms with a constant and not with each other, i.e., x/a=k, y/b=k and z/c=k, therefore x/a=y/b=z/c. Is there any way by which we can do this?

Answer 1

It is tacitly assumed that $abc\ne0$. The claim suggests to replace the variables $x$, $y$, $z$ by new variables $\xi$, $\eta$, $\zeta$ as follows: $$x=a\xi,\quad y=b\eta,\quad z=c\zeta\ .$$ Then your equations say $${ab\over c}(\eta-\xi)={ac\over b}(\xi-\zeta)={bc\over a}(\zeta-\eta)\ .$$ Multiply by $abc$ and obtain $$a^2b^2(\eta-\xi)=a^2c^2(\xi-\zeta)=b^2c^2(\zeta-\eta)\ .\tag{1}$$ This implies that the three real numbers $\eta-\xi$, $\xi-\zeta$, $\zeta-\eta$, if nonzero, all have the same sign. Since they sum to zero it follows that in fact $\xi=\eta=\zeta$.

Note that the claim is wrong if $a$, $b$, $c$ are allowed to be complex: If $a^2+b^2+c^2=0$ then the given equations collapse to the single equation $ax+by+cz=0$ allowing of additional solutions. As an example, consider $a=b=1$, $c=i\sqrt{2}$. Then $(1)$ becomes $$\eta-\xi=-2(\xi-\zeta)=-2(\zeta-\eta)$$ which is equivalent to the single equation $\xi+\eta-2\zeta=0$. This is not only solved by $(\xi,\eta,\zeta)=\lambda(1,1,1)$, but has a two-dimensional solution space.