在正方形ABCD中,BE⊥AP,EF=AE,BF=BC,∠CBF的角平分线交AF于G,(1)证BG+DG=根号下2AG (2)AB=2,P为BC中点,CF
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![在正方形ABCD中,BE⊥AP,EF=AE,BF=BC,∠CBF的角平分线交AF于G,(1)证BG+DG=根号下2AG (2)AB=2,P为BC中点,CF](/uploads/image/z/12086216-8-6.jpg?t=%E5%9C%A8%E6%AD%A3%E6%96%B9%E5%BD%A2ABCD%E4%B8%AD%2CBE%E2%8A%A5AP%2CEF%3DAE%2CBF%3DBC%2C%E2%88%A0CBF%E7%9A%84%E8%A7%92%E5%B9%B3%E5%88%86%E7%BA%BF%E4%BA%A4AF%E4%BA%8EG%2C%281%29%E8%AF%81BG%2BDG%3D%E6%A0%B9%E5%8F%B7%E4%B8%8B2AG+%282%29AB%3D2%2CP%E4%B8%BABC%E4%B8%AD%E7%82%B9%2CCF)
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