（1）∵∠BAD=∠ECD,∠ABD=∠CED,∴△ABD∽△CED,∴ CD:AD=CE:AB,∴CD=3．证明：（2）连接IB．∵点I是△ABC的内心,∴∠BAD=∠CAD,∠ABI=∠CBI,∴弧BE=弧CE,则BE=CE,∴∠BIE=∠BAD+∠ABI=∠IBD+∠CAD=∠IBD+∠CBE=∠I...