证明:作直径AF,连接BF,作OG⊥AB 于点G
∵AF是直径
∴∠ABF=90°
∵∠F=∠C,∠ADC=90°
∴∠OAG=∠HAE
∵∠OGA=∠HEA=90°,AO=AH
∴△OAG≌△HAE
∴AE=AG
∵OG⊥AB
∴AG=1/2AB
∴AE =1/2AB
∵∠AEB =90°
∴∠ABE =30°
∴∠ABC =60°
证明:作直径AF,连接BF,作OG⊥AB 于点G
∵AF是直径
∴∠ABF=90°
∵∠F=∠C,∠ADC=90°
∴∠OAG=∠HAE
∵∠OGA=∠HEA=90°,AO=AH
∴△OAG≌△HAE
∴AE=AG
∵OG⊥AB
∴AG=1/2AB
∴AE =1/2AB
∵∠AEB =90°
∴∠ABE =30°
∴∠ABC =60°