Question

如图,已知等边三角形ABC中,点E是射线BC上的点,点F是射线CA上的点,且BE=CF与直线AE交于点P,就下面给出的三

Answer 1

∠BPE=60按图3证明：AB=CA ∠BAF=∠ACE=180-60=120AF=CF-CA CE=BE-BC CF=BE CA=BCAF=CE △BAF≌△ACE∠FBA=∠EAC ∠FAP=∠EAC ∠FAP=∠FBA∠BPE=∠F+∠FAP ∠BAC=∠F+∠FBA∠BPE=∠BAC=60...