Question

如图，三角形ABC中，AD垂直BC于点D，AD=DC，点F在AD上，角FCD=角BAD，BF的延长线交AC于点E.求证：BE垂直A

如图，三角形ABC中，AD垂直BC于点D，AD=DC，点F在AD上，角FCD=角BAD，BF的延长线交AC于点E.求证：BE垂直AC；设CE的长为m，用含m的代数式表示AC+BF

Answer 1

AD=DC，∠FCD=∠BAD
∠FDC=90, ∠BDF=90
△ABD , △CFD全等

BD=FD.∠DBF = ∠BFD= 45
AD=DC, ∠CAD = ∠ACD=45
BEC = 180 -∠DBF - ∠ACD=90
所以 BE⊥AC

2
EBC=45, ECB =45,
BE =CE =m
AFE=BFD= 45
FAE =45
所以 AE= EF
AC = AE+CE
BF = BE-EF
AC+BF = AE+CE + BE-EF = CE + BE= 2m