This article gives a proof of Ceva's theorem by using Menelaus's theorem twice. However, it also states:
Oddly, it’s somewhat difficult to go the other way. To prove Menelaus from Ceva requires using Ceva six times!
I tried this but didn't get a perfect result:
(All segments are directed, i.e. $AB=-BA$.)
Apply Ceva onto $\triangle ABE$ and $D$: $$\frac{AF}{FB}\cdot\frac{BG}{GE}\cdot\frac{EC}{CA}=1$$ Apply Ceva onto $\triangle BDE$ and $A$: $$\frac{BC}{CD}\cdot\frac{DF}{FE}\cdot\frac{EG}{GB}=1$$ Apply Ceva onto $\triangle BEF$ and $C$: $$\frac{BH}{HE}\cdot\frac{ED}{DF}\cdot\frac{FA}{AB}=1$$ Apply Ceva onto $\triangle BCE$ and $F$: $$\frac{BD}{DC}\cdot\frac{CA}{AE}\cdot\frac{EH}{HB}=1$$ Apply Ceva onto $\triangle AFC$ and $D$: $$\frac{AB}{BF}\cdot\frac{FJ}{JC}\cdot\frac{CE}{EA}=1$$ Apply Ceva onto $\triangle CFD$ and $A$: $$\frac{CJ}{JF}\cdot\frac{FE}{ED}\cdot\frac{DB}{BC}=1$$
By multiplying these 6 equations, I got: $$\frac{AF^2}{FB^2}\cdot\frac{BD^2}{DC^2}\cdot\frac{CE^2}{EA^2}=1$$
which is very close to Menelaus's theorem but the direction or sign is not determined.
Is there any improvement or I'm in a wrong way?