(In reply to

I think I got it. by Jer)

Nice Proof.

But there is a much simpler solution that covers the more general case where each of a, b and c are any nonzero real numbers. (The specific reason of stating that each of a, b and c is positive is to keep the denominator of each of the three expressions nonzero.)

HINT

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Assume that (ay-bx)/c = (cx-az)/b = (bz-cy)/a =p(say)

Then arrive at:

A quadratic expression involving a, b and c multiplied by something is equal to zero and, everything else will just fall in place.

Spoiler Alert:

The quadratic expression and that something are much more simpler than envisaged.

*Edited on ***November 25, 2012, 3:36 am**