1. ## Problemo

I need help to solve this problem.

If $\displaystyle (ay - bx):c = (cx - az):b = (bz - cy):a$ then prove that
$\displaystyle \frac{x}{a} = \frac{y}{b} = \frac{z}{c}$.

$\displaystyle a,b,c \ne 0$

2. Hello, OReilly!

Here's a start . . .

(But I sincerely hope there's a better method!)

If . $\displaystyle \frac{ay - bx}{c} \,= \,\frac{cx - az}{b} \,= \,\frac{bz - cy}{a}$, then prove that: .$\displaystyle \frac{x}{a} = \frac{y}{b} = \frac{z}{c}\;\;\;(a,b,c \ne 0)$
. . . . .[1] . . . . . . . [2] . . . . . . .[3]

From [1] = [3], we have: .$\displaystyle \frac{ay-bx}{c} \:= \:\frac{bz-cy}{a}\quad\Rightarrow\quad a^2y$$\displaystyle -abx \:=\:bcz - c^2y . . Solve for \displaystyle x:\;\;x\:=\:\frac{a^2y + c^2y - bcz}{ab} . (a) From [2] = [3], we have: .\displaystyle \frac{cx-az}{b} \:= \:\frac{bz-cy}{a}\quad\Rightarrow\quad acx$$\displaystyle -a^2z \:= \:b^2z - bcy$

. . Solve for $\displaystyle x:\;\;x\:=\:\frac{a^2z + b^2z - bcy}{ac}$ .(b)

Equate (a) and (b): .$\displaystyle \frac{a^2y + c^2y - bcz}{ab} \:= \:\frac{a^2z + b^2z - bcy}{ac}$

. . and we have: .$\displaystyle a^2cy + c^3y - bc^2z \:= \:a^2bz + b^3z - b^2cy$

Rearrange terms: .$\displaystyle a^2cy + b^2cy + c^3y \:= \:a^2bz + b^3z + bc^2z$

Factor: .$\displaystyle cy(a^2+b^2+c^2) \:=\: bz(a^2+b^2+c^2)$

Divide by $\displaystyle (a^2+b^2+c^2):\;\;cy\,= \,bz\quad\Rightarrow\quad\boxed{ \frac{y}{b} \,= \,\frac{z}{c} }$

I'll let you work on the other equality . . .

3. Ye, not very ellegant solution, but it works.
Thanks!