如图,已知AC为正方形ABCD的对角线,点P为AC上任意一点,过点P作PE⊥BP交CD于点E,BE交AC于点F.当AP：

作∠ABG＝∠CBF,使G、F在AB的两侧且BG＝BF.令AP＝4t、CF＝3t.
∵ABCD是正方形,∴DC＝AB＝CB、BC⊥CE、∠ABC＝90°、∠BAP＝∠ACD＝∠BCF＝45°.
∵BC⊥CE,又PE⊥BP,∴B、C、E、P共圆,∴∠PBE＝∠ACD＝45°.
∵∠ABC＝90°、∠PBE＝45°,∴∠CBE＋∠ABP＝∠ABC－∠PBE＝45°,而∠ABG＝∠CBE,
∴∠ABG＋∠ABP＝45°,∴∠PBG＝45°.
∵AB＝CB、BG＝BF、∠ABG＝∠CBF,∴△ABG≌△CBF,∴AG＝CF、∠BAG＝∠BCF＝45°.
∵∠BAP＝∠BAG＝45°,∴∠PAG＝90°,又AP＝4t、CF＝3t,∴PG＝5t.
∵BG＝BF、BP＝BP、∠PBG＝∠PBF＝45°,∴△PBG≌△PBF,∴PG＝PF＝5t.
∵ABCD是正方形,∴CE∥AB,∴CE/AB＝CF/AF＝CF/（AP＋PF）＝3t/（4t＋5t）＝1/3,
∴CE/DC＝1/3,∴CE/（ED＋CE）＝1/3,∴CE/ED＝1/（3－1）＝1/2,∴ED＝2CE.