Question

如图,在矩形ABCD中,AC,BD交于点O,E是CB延长线一点,CF⊥AE,垂足为F,请说明:DF⊥BF

Answer 1

连接OF,

∵四边形ABCD是矩形 ∴OA=OC=OB=OD

∵CF⊥AE ∴∠AFC=90° ∴OF=½AC=OA ∴OF=OB=OD

∴∠ODF=∠OFD ,∠OFB=∠OBF

∵⊿BDF中 ∠ODF＋∠DFB＋∠OBF=180°即∠ODF＋∠OFD＋∠OFB＋∠OBF=180°

∴2﹙∠OFD＋∠OFB﹚=180° ∴∠OFD＋∠OFB=90°即∠DFB=90°

∴DF⊥BF