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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 $FH:DH$.

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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.

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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?

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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.

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