∵EF⊥BC,AD⊥BC∴EF∥AG(AD)∵CE平分∠ACB,即∠ACE=∠FCE(∠BCE)∠EAC=∠EFC=90°(∠BAC=∠EFC=90°)CE=CE∴△ACE≌△FCE(AAS)∴AE=EF∵∠AEG=90°-∠ACE∠DGC=90°-∠BCE(∠FCE)∴∠AEG=∠DGC=∠AGE(对顶角)∴AE=AG=EF∵EF=AG,EF∥AG∴AEFG是平行四边形∵AE=EF=AG∴AEFG是菱形