∵AC=BC,∴∠CAD=∠CBD.
∵A、C、B、E共圆,∴∠CBD=∠AEC,∴∠CAD=∠AEC,又∠ACD=∠ECA,
∴△ACD∽△ECA,∴∠ADC=∠EAC.
∵CE是⊙O的直径,∴∠EAC=90°,∴∠ADC=90°.
由AC=BC、CD⊥AB,得:AD=AB/2=2.
∵AC⊥AE、AD⊥CE,∴由射影定理,有:CD×DE=AD^2=4,∴DE=4/CD=4.
∴CE=CD+DE=1+4=5,∴OC=CE/2=5/2.
∴⊙O的半径是5/2.
∵AC=BC,∴∠CAD=∠CBD.
∵A、C、B、E共圆,∴∠CBD=∠AEC,∴∠CAD=∠AEC,又∠ACD=∠ECA,
∴△ACD∽△ECA,∴∠ADC=∠EAC.
∵CE是⊙O的直径,∴∠EAC=90°,∴∠ADC=90°.
由AC=BC、CD⊥AB,得:AD=AB/2=2.
∵AC⊥AE、AD⊥CE,∴由射影定理,有:CD×DE=AD^2=4,∴DE=4/CD=4.
∴CE=CD+DE=1+4=5,∴OC=CE/2=5/2.
∴⊙O的半径是5/2.