几何
已知:如图,矩形ABCD中,AC与BD交于点O,AF⊥BD,垂足为F,∠BCD的平分线交FA的延长线于点E。求证:AE=AC
∠ACE =∠BCD/2 -∠ACD =45°-∠ACD ∠AEC =∠CAF-∠ACE =(90°-∠AOF)-∠ACE = (90°-2*∠ABD)-(45°-∠ACD) = 45°-2*∠ABD+∠ACD =45°-∠ACD =∠ACE ===> AE=AC
答:猜想 EF=PD 延长EP交CD于G,延长FP交AD于点H。 由于点P在正方形的对角线AC上,有FH⊥AD,EG⊥CD,且PF=PG=HD,PE=PH。由此可证...详情>>
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