Question

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

Answer 1

∠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