Let $M,N,H$ be the midpoints of $AD,AB,BC$ respectively. Let $k_1 , k_2$ be the circles with diameters $AD,AB$ respectively. Let $G = k_1\cap k_2 \neq A$. It is easy to see that $B,G,D$ are collinear. Also $N_3 N_4\parallel BC , N_2 N_3\parallel AB, N_1 N_4\parallel CD , N_1 N_2\parallel AD$ so by a homothety we have $N_1 N_2 N_3 N_4$ are concyclic. Also it is not too hard too see that the radius of $(N_1 N_2 N_3 N_4)$ is half the radius of $(ABCD)$. Also $\angle FGA = \angle FBA = 180 - \angle ABC = \angle ADC = 180 - \angle EGA \implies F,G,E$ are collinear. Let $K$ be the center of $(N_1 N_2 N_3 N_4)$. We have $N,K,H,F$ and $N,K,G,M$ are concyclic. So $\angle NKF = \angle NHF = \angle NHB = \angle ACB = \angle ADB = \angle ADG = \angle NMG = 180 - \angle NKG = 180 - \angle NKE \implies E,K,F$ are collinear. So $EF$ bisects the circle $(N_1 N_2 N_3 N_4)$.