Parallel lines

Since $\angle BEC \cong \angle HEA$ and $\angle HAC \cong \angle ACB$ , $\triangle HEA \sim \triangle BEC$ (A.A.) and $\dfrac{AE}{EF+CF}=\dfrac{HE}{BE}$

Similarly, $\triangle BFA \sim \triangle GFC$ and $\dfrac{CF}{EF+AE}=\dfrac{FG}{BF}$

However, since $CF=AE$ , $\dfrac{HE}{BE}=\dfrac{FG}{BF}$ .

Therefore, in $\triangle HBG$ , $HG \parallel EF$ or $HG \parallel AC$