The most difficult thing is the second question. I think the correct answer is that DEG is a regular triangle, ∠DAC=∠GHI=∠EFB, and those nine lines are basically unnecessary to be parallel, otherwise they are all regular triangles. I tried it with CAD drawing, and it completely conforms to the conclusion, so we can get the conclusion we want without parallelism. Because DEG is a regular triangle, we get ∠ feb = ∠ ADC = ∠ HGI = 60. ∠DAC=∠GHI=∠EFB, according to AAS, we get △ ACD △ BEF △ GED △ GHI, so we get GH=EF=AD.

GI=EB=CD, with three identical sides of DEG on each side. HE=FD=AG, DI=GB=CE, because ∠GDE=∠DEG=∠DGE, △ ABG △ Che △ IFD is obtained according to SAS.