Question

点E在正方形ABCD的边CB的延长线上,DE与边AB相交与点F,FG‖BE,FG与AE相交与点G.求证：GF=BF

Answer 1

直角三角形EBF与直角三角形ECD相似:

EB:EC=BF:CD,

EB:(EB+BC)=BF:BC,

BF=EB*BC/(EB+BC),

AF=AB-BF=BC-EB*BC/(EB+BC)=(BC)²/(EB+BC);

直角三角形AFG与直角三角形ABE相似:

GF:EB=AF:AB=AF:BC,

GF=EB*AF/BC=EB*(BC)²/(EB+BC)/(BC)=EB*BC/(EB+BC)=BF.