见解析
证明 (1)如图③,连接BE.
∵AB=AC,∴∠ABC=∠ACB.
∵∠ACB=∠AEB,
∴∠ABC=∠AEB.
∴△ABD∽△AEB.
∴AB∶AE=AD∶AB,
即AB 2=AD·AE.
(2)如图④,连接BE、EC,
∵四边形ABCE内接于⊙O,
∴∠CED=∠ABC,
∵AB=AC,∴∠ABC=∠ACB,
∴∠CED=∠ACB,
∵∠AEC=180°-∠CED,
∠ACD=180°-∠ACB,
∴∠AEC=∠ACD,∴△ACE∽△ADC,
∴=,∴AB 2=AD·AE.
见解析
证明 (1)如图③,连接BE.
∵AB=AC,∴∠ABC=∠ACB.
∵∠ACB=∠AEB,
∴∠ABC=∠AEB.
∴△ABD∽△AEB.
∴AB∶AE=AD∶AB,
即AB 2=AD·AE.
(2)如图④,连接BE、EC,
∵四边形ABCE内接于⊙O,
∴∠CED=∠ABC,
∵AB=AC,∴∠ABC=∠ACB,
∴∠CED=∠ACB,
∵∠AEC=180°-∠CED,
∠ACD=180°-∠ACB,
∴∠AEC=∠ACD,∴△ACE∽△ADC,
∴=,∴AB 2=AD·AE.