过矩形ABCD的顶点C引BD的垂线与角BAD的平分线交于点E,垂足为F,求证：BD=CE

过矩形ABCD的顶点C引BD的垂线与角BAD的平分线交于点E,垂足为F,求证：BD=CE

证明:设AE交BC于M.
AE平分∠BAD,则∠BAE=45°=∠BMA=∠CME.
连接AC,四边形ABCD为矩形,则AC=BD;∠BAC=∠CDB=∠BAM+∠CAE=45°+∠CAE.--------(1)
又CF垂直BD,则∠CDB=∠BCF(均为∠DCF的余角).
故∠CDB=∠BCF=∠CME+∠CEA=45°+∠CEA.------------------------------------------------------(2)
∴∠CAE=∠CEA,得:AC=CE,故BD=CE.