(1)证明:∵ | CD | = | BD | , ∴CD=BD, ∵∠CDB=60°, ∴△BCD是等边三角形, ∴ | CD | = | BC | , ∴∠CAD=∠BAC,即AC是∠DAB的平分线;
(2)连接BD,在线段CE上取点F,使得EF=AE,连接DF, ∵DE⊥AC, ∴DF=DA, ∴∠DFE=∠DAE, ∵ | CD | = | BD | , ∴CD=BD,∠DAC=∠DCB, ∴∠DFE=∠DCB, ∵四边形ABCD是圆的内接四边形, ∴∠DAB+∠DCB=180°, ∵∠DFC+∠DFE=180°, ∴∠DFC=∠DAB, ∵在△CDF和△BDA中,
∴△CDF≌△BDA(AAS), ∴CF=AB=5, ∵AC=7,AB=5, ∴AE=AF=(AC-CF)=1. |