Let $ABC$ be a triangle. $D, E, F, G, H$ are points such as $E~\in AC$, $D ~\in BC$, $F = AD\cap BE$, $G = \overleftrightarrow{CF} \cap AB$ and $H = ED \cap CG$.
Prove that $\dfrac{HF}{HC} = \dfrac{FG}{CG}$
I'm pretty sure we should use Menelaus's theorem or Ceva's theorem (maybe a combination of both). But I tried a lot of combinations and no one were able to generate what we want.
Here's a draft of the ones I wrote down and didn't work:
Ceva's: CE * AG * BD / EA * BG * CD = 1
Menelaus' AGC: GB * EA * CF/ AB * EC * GF = 1
Menelaus' AFC: CE * HF * AD/ EA * HC *DF = 1
Menelaus' ABE: EC * GA * FB/ CA * GB * FE = 1
Menelaus' BCG: DB * FC * AG/ DC * FG * AB = 1
Menelaus' BCE: DB * FE * AC/ DC * FB * AE = 1
Menelaus' CEF: AC * GF * BE / AE * GC * BF= 1
Menelaus' CDF: AD * GF * BC / AF * GC * BD = 1
Menelaus' BFC: EB * HF * DC / EF * HC * DB = 1
Menelaus' AFG: CF * DA * BG / CG * DF * BA = 1
Menelaus' CDH: EH * FC * BD / ED * FH * BC = 1
Menelaus' CHE: DE * FH * AC / DH * FC * AE = 1
Somebody help me, please!
We need to prove that $[FC,HG]=-1.$
Indeed, Let $ED\cap AB=\{N\}$.
Thus, we have $ECDF$ is a complete quadrangle. See here: https://en.wikipedia.org/wiki/Projective_harmonic_conjugate
Thus, $[AB,GN]=-1.$
Now, let $\phi$ be a projection of $CG$ on $AB$ from a point $E$.
Thus, $$\phi(F)=B,$$$$\phi(C)=A,$$ $$\phi(H)=N$$ and $$\phi(G)=G.$$ Id est, $$[FC,HG]=[BA,NG]=[AB,NG]=[AB,GN]=-1$$ and we are done!