解:(1)∵四边形ABCD内接于⊙O,∴∠CDA=∠ABE ∵,∴∠DCA=∠BAE ∴△CAD∽△AEB (2)过A作AH⊥BC于H ∵A是中点,∴HC=HB=BC ∵∠CAE=90°,∴AC2=CH·CE=BC·CE (3)∵A是中点,AB=2,∴AC=AB=2, ∵EM是⊙O的切线 ∴EB·EC=EM2 ① ∵AC2=BC·CE,BC·CE=8 ② ①+②得:EC(EB+BC)=17,∴EC2=17 ∵EC2=AC2+AE2,∴AE= ∵△CAD∽△ABE,∴∠CAD=∠AEC ∴cot∠CAD=cot∠AEC= |