S△ABC=S△AEG
证明:
作BM⊥AC于点M,EN⊥AG,交GA的延长线于点N
则∠EAN+∠BAN=∠MAB+∠BAN=90°
∴∠EAN=∠BAM
∵∠ENA=∠AMB=90°,AE=AB
∴△EAN≌△BAM
∴EN=BM
∵S△ABC=1/2*AC*BM,S△AEG=1/2*AG*EN,AC=AG,BM=EN
∴S△ABC=S△AEG
S△ABC=S△AEG
证明:
作BM⊥AC于点M,EN⊥AG,交GA的延长线于点N
则∠EAN+∠BAN=∠MAB+∠BAN=90°
∴∠EAN=∠BAM
∵∠ENA=∠AMB=90°,AE=AB
∴△EAN≌△BAM
∴EN=BM
∵S△ABC=1/2*AC*BM,S△AEG=1/2*AG*EN,AC=AG,BM=EN
∴S△ABC=S△AEG