过点D作DE⊥AC于点C,DF⊥AB于点F
∵∠BAC=90º,AD平分∠BAC
∴四边形AEDF是正方形
设AE=AF=DE=DF=x
∵ΔDEC∽ΔBFD
∴x/(3-x)=(2-x)/x
解得x=4/3
即AE=4/3
∴AD=4√2/3