矩形ABCD中,CE垂直于BD于E,AF平分角BAD交EC的延长线于F,交BC于G,交BD于H,求:CA=CF
矩形ABCD中,CE垂直于BD于E,AF平分角BAD交EC的延长线于F,交BC于G,交BD于H,求:CA=CF
数学人气:652 ℃时间:2019-12-13 01:23:02
优质解答
只需证明角CAF=角F,设AC,BD交于点O,角BHF=角AOD+角CAF,角AOD=2倍的角BAC,那么角BHF=2倍的角BAC+角CAF=2倍的角BAF-角CAF=90度-角CAF,那么角CAF=90度-角BHF=角F,所以CA=CF
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