解:∵CA是∠BCD的平分线,∴∠1=∠2,∵AD∥BC,∴∠2=∠3,从而∠1=∠3,∵AD=6,∴CD=AD=6,过点D作DE⊥AC于E,则AE=CE= 1 2 AC,∵∠1=∠2,∠BAC=∠DEC,∴△ABC∽△EDC,∴ CD BC = CE AC ,即 6 BC = 1 2 ,∴BC=12,在Rt△ABC中,由勾股定理得,AC= BC2?AB2 = 122?42 =8 2 .