题目内容：

正方形ABCD中,P 是对角线AC上一动点,PE⊥AB,PF⊥BC,垂足分别为E、F求证DP=EF DP⊥EF

优质解答

延长FP交AD于点Q,连接BP.则BEPF是矩形,可得：EF = BP ,∠EFP = ∠EBP ；因为,在△ADP和△ABP中,AD = AB ,∠DAP = ∠BAP ,AP为公共边,所以,△ADP ≌ △ABP ,可得：∠ADP = ∠ABP = ∠EFP ,DP = BP = EF ；延长DP交EF...