(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
Scroll Down
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