摘要： 如图9.点P是正方形ABCD边AB上一点(不与点A.B重合).连接PD并将线段PD绕点P顺时针方向旋转90°得到线段PE.PE交边BC于点F.连接BE.DF. (1)求证:∠ADP＝∠EPB, (2)求∠CBE的度数, (3)当的值等于多少时.△PFD∽△BFP?并说明理由. [答案] (1)证明:∵四边形ABCD是正方形 ∴∠A＝∠PBC＝90°.AB＝AD.∴∠ADP+∠APD＝90°················ 1分 ∵∠DPE＝90° ∴∠APD+∠EPB＝90° ∴∠ADP＝∠EPB.········································································································ 2分 (2)过点E作EG⊥AB交AB的延长线于点G.则∠EGP＝∠A＝90°·· 3分 又∵∠ADP＝∠EPB.PD＝PE.∴△PAD≌△EGP ∴EG＝AP.AD＝AB＝PG.∴AP＝EG＝BG················································· 4分 ∴∠CBE＝∠EBG＝45°.························································································· 5分 (3)方法一: 当时.△PFE∽△BFP.·············································································· 6分 ∵∠ADP＝∠FPB.∠A＝∠PBF.∴△ADP∽△BPF······························ 7分 设AD＝AB＝a.则AP＝PB＝.∴BF＝BP····················· 8分 ∴. ∴··········································································································· 9分 又∵∠DPF＝∠PBF＝90°.∴△ADP∽△BFP·········································· 10分 方法二: 假设△ADP∽△BFP.则.·································································· 6分 ∵∠ADP＝∠FPB.∠A＝∠PBF.∴△ADP∽△BPF··························· 7分 ∴.··············································································································· 8分 ∴.··············································································································· 9分 ∴PB＝AP. ∴当时.△PFE∽△BFP. 10分

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