Question

在矩形ABCD中,AC,BD交于O点H是AD一点（H与A,D不重合）HE⊥AC交于E,HF⊥BD交于F,AG⊥BD交于G

Answer 1

证明：连接OH

∵矩形ABCD

∴AO＝DO

∵HE⊥AC,HF⊥BD

∴S△AOH＝AO×HE/2,S△DOH＝DO×HF/2

∵AG⊥BD

∴S△AOD＝DO×AG/2

∵S△AOH+S△DOH＝S△AOD

∴AO×HE/2+DO×HF/2＝DO×AG/2

∴HE+HF=AG