如图,已知△ABC中,∠BAC=90°,AB=AC,点P为BC边上一动点(BP<CP),分别过B、C作BE⊥AP于E,C
如图,已知△ABC中,∠BAC=90°,AB=AC,点P为BC边上一动点(BP<CP),分别过B、C作BE⊥AP于E,CF⊥AP于F.
求证:①△ABE≌△CAF;②EF=CF-BE
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问答/359℃/2024-04-27 08:50:35
优质解答:
∠E = ∠AFC = 90°
(1)∠BAE+∠EAC=90°∠BAE+∠ABE =90°所以∠ABE=∠EAC 同理∠BAE=∠ACF
(2)AB=AC △ABE≌△CAF (AAS)
AE-AF=EF AE = CF AF=BE 所以CF-BE =EF