如图,在锐角△ABC中,D,E分别是AB,AC边上的点,△ADC≌△ADC′,△AEB≌△AEB′,且C′D∥EB′∥BC,BE,CD交于点F,若∠BAC=40°,则∠BFC的度数是( )

- A.105°
- B.100°
- C.110°
- D.115°
答案
正确答案:B

延长C′D交AB′于H.
∵△AEB≌△AEB′,
∴∠ABE=∠AB′E,
∵C′H∥EB′,
∴∠AHC′=∠AB′E,
∴∠ABE=∠AHC′,
∵△ADC≌△ADC′,
∴∠C′=∠ACD,
∵∠BFC=∠DBF+∠BDF,∠BDF=∠CAD+∠ACD,
∴∠BFC=∠AHC′+∠C′+∠DAC,
∵∠DAC=∠DAC′=∠CAB′=40°,
∴∠C′AH=120°,
∴∠C′+∠AHC′=60°,
∴∠BFC=60°+40°=100°,

略
