If $\frac{ay-bx}{p}=\frac{cx-az}{q}=\frac{bz-cy}{r} $

Question

If $$\frac{ay-bx}{p}=\frac{cx-az}{q}=\frac{bz-cy}{r}$$, prove that : $$\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$$

My solution, $$\frac{c(ay-bx)}{cp}=\frac{b(cx-az)}{bq}=\frac{a(bz-cy)}{ar}$$. Now, I could not get enough idea to continue from herr.

Answer 1

In your last equation, call the ratio $k$, so that $$\left\{\begin{array}{}c(ay-bx)& = & kcp\\ b(cx-az) & = & kbq\\ a(bz-cy) & = & kar\end{array}\right.$$ Taking the sum of all these equations yields : $$acy-bcx+bcx-abz+abz-acy=k(cp+bq+ar).$$The left-hand side is $0$, and $(cp+bq+ar)$ is positive, thus $k=0$. It follows that $$ay-bx=cx-az=bz-cy=0,$$and thus $$\frac{x}{a}=\frac{y}{b}=\frac{z}{c}.$$

Answer 2

$$\frac{ay-bx}p=\frac{\dfrac yb-\dfrac xa}{abp}$$

If $\dfrac yb=\dfrac xa,\dfrac{ay-bx}p=0$

So, each of the given ratio has to be zero