如图,AB//CD,BE平分角ABF,DE平分角CDF,角BFD=120度,求角BOD.

Question

如图,AB//CD,BE平分角ABF,DE平分角CDF,角BFD=120度,求角BOD.

Answer 1

连接BD,在三角形BFD中,因为∠BFD=120°,所以∠DBF+∠BDF=180°-120°=60°
因为BE平分∠ABF,DE平分∠CDF
所以∠FBE=1/2∠ABF ∠FDE=1/2∠CDF
AB平行CD,所以∠ABD+∠CDB=180°
在三角形BED中,∠BED=180-(∠EBD+∠EDB)
=180-(∠FBE+∠DBF+∠FDE+∠BDF )
=180-(1/2∠ABF+1/2∠CDF+60)
=180-1/2(∠ABF+∠CDF+2∠FDE+2∠BDF)
=180-1/2(180+60)
=60