If (ay-bx)/c=(cx-az)/b=(bz-cy)/a, then how to prove that x/a=y/b=z/c?

1 Answer
Jul 6, 2017

(ay-bx)/c=(cx-az)/b=(bz-cy)/a

"each"= (acy-bcx)/c^2=(bcx-abz)/b^2=(abz-cay)/a^2

By addendo we get

"each"=( (acy-bcx)+(bcx-abz)+(abz-cay))/(a^2+b^2+c^2)=0

So (ay-bx)/c=0=>x/a=y/b

(cx-az)/b=0=>x/a=z/c

Hence

x/a=y/b=z/c