Question

已知正方形ABCD的BC边上有一点F,BC边上的延长线为CE.CM是∠DCE的平分线,∠AFG=90°,FG交CM于点G

Answer 1

AF⊥GF

∠BAF+∠AFB=∠AFB+∠GFH=90°

∠BAF=∠GFH

tan∠BAF=BF/AB=BF/(BF+FC)

CG平分∠DCH

∠DCG=∠GCH=∠CGH=45°

GH=CH

tan∠GFH=GH/FH=GH/(GH+FC)

BF/(BF+FC)=GH/(GH+FC)

所以BF=GH