Answer:
(i) ABCD = ABEC
(ii) As BF || AC then BE || AC, so CE = AB.
As AB = DC, then DC = CE
(iii) ΔDGE ≠ ΔCFE so I assume there is a typing error in the question, as
ΔDGC = ΔCFE
As DC = CE (as proved in (i)), AC || BE and DG ⊥ AC and CF ⊥ BE then ΔDGC = ΔCFE
(iv) As DG ⊥ AC and CF ⊥ BE, and AC || BE, then DG || CF
(v) As AC || BE then GF || DC. As DG || CF (as proved in (iv)), then DGFC is a parallelogram.