(1)①证明:∵∠ACB=90°, ∴∠ACE+∠BCD=180°﹣90°=90°, ∵BD⊥DE,AE⊥DE, ∴∠E=∠D=90°, 在△ACE中,∠ACE+∠CAE=180°﹣∠E=90°, ∴∠CAE=∠BCD, ∴△ACE∽△CBD; ②解:∵△ACE∽△CBD;AC=BC, ∴△ACE≌△CBD, ∵AE=a,BD=b, ∴ ×CE=b,CD=a, ∵ED为梯形AEDB的高, ∴△ACB的面积=梯形AEDB面积﹣△ACE面积﹣△CBD的面积, = ×DE﹣ AE·EC﹣ BD·DC, = ×(a+b)﹣ ab﹣ ab, = (a2+b2), (2)解:∵△ACE≌△CBD(已证), ∴CD=AE,CE=BD, ∵AE=a,BD=b, ∴CE=b,CD=a, ∴DE=CE﹣CD=b﹣a, ∴梯形ADBE面积= ×DE, = ×(b﹣a), = . |