Proof: Points F and E are parallel to CD as FP and EM respectively, and submitted to BH and BG respectively, which can be proved.

FP:CH=BF:BC=2:3, EM:CG=BE:BC= 1:3, so FP=2CH/3, ME=CG/3.

Because CH=CD/3, CG=2CD/3 and CD=AB, FP=2CD/9=2AB/9, ME=2AB/9.

Press FP‖CD, ME‖CD , AB‖CD, so? FP‖AB，？ ME‖AB, so it can be proved.

AN:FN=AB:FP=AB:2AB/9=9:2, AK:KE=AB:ME=AB:2AB/9=9:2, so

AN:AF=9: 1 1, AK:AF=9: 1 1, so AN:AF=AK:AF and ∠KAN=∠EAF, so

△AKN∽△AEF, so ∠ANK=∠AFE, so KN ∠ EF.