In $\triangle ABC$, $D, E, F, G$ are points on the sides of the triangle such that $BD:DE:EC=1:2:3$, $AF:CF=1:1$, and $AG:BG=2:3$. Find the ratio $FH:DH$.
My classmate has come up with a brilliant way to do this. He argued that if we let $$AC=1, AB=5,BC=6,$$
The triangle will collapse into a straight line.
We can observe directly
$$FH:DH=2.5:2=5:4$$
Despite the lack of rigor, this method successfully computes the right solution. My question is:
Can we transform the limiting case method (as I would call it) to a rigorous answer?
Yes, for this case, basically it's equivalent to projecting everything onto $BC$ along direction of the line $GHE$. Since we're looking for a ratio, it doesn't change when we perform projection.