（1）证明：由直棱柱的性质可得，AA1⊥平面ABC∴AA1⊥AB∵在△ABC中AB=1，AC=3，BC=2，AB2+AC2=BC2∴AB⊥AC又AC∩AA1=A∴AB⊥平面ACC1A1，又∵A1C⊂平面ACC1A1∴AB⊥A1C（2）连接A1C，A1B由已知可得A1B＝BC＝2 ...