取BC中点K
连接MK、KN
KM‖CE,KM=1/2CE
∠AHG =∠NMK
KN ‖BD,KN=1/2BD
∠AGH=1/2∠MNK
KM=KN
∠NMK=∠MNK
∠AHG=∠AGH
AG=AH