如图,点D、E分别在△ABC的边AB、AC上,且AB=9,AC=6,AD=3,若△ADE与△ABC相似,则AE的长为()
①当△ADE∽△ACB时,$\frac {A E} {A B} = \frac {A D} {A C}$,即$\frac {A E} {9} = \frac {3} {6}$,解得AE=$\frac {9} {2}$;②当△ADE∽△ABC时,$\frac {A D} {A B} = \frac {A E} {A C}$,即$\frac {3} {9} = \frac {A E} {6}$,解得AE=2. 故选选项C.