解:(1)∵BD=BC, ∴∠DCB=∠D. -----------------------------------(1分) 又∵CE⊥CD,∠ACB=, ∴∠DCB+∠BCE=, ∠ACE+∠BCE=, ∴∠D=∠DCB=∠ACE,-----------------------------(2分) 又∵∠A =∠A ,-----------------------------------(1分) ∴△ACE∽△ADC. --------------------------------(1分) (2)∵∠DCB+∠BCE=, ∠D+∠DEC=,又∠DCB=∠D, ∴∠BCE=∠BEC,-----------------------------------(1分) ∴BE="BC." ----------------------------------------(1分) 又BE∶EA=3∶2,令BE=3k,EA="2" k, ----------------(1分) 在△ABC中,∠ACB=,BC=3k,AB=5k,-----------(1分) ∴sin∠A=.---------------------------------(1分) |