已知:如图,设M是△ABC内部任意一点,MD⊥AB于G,ME⊥BC于K,MF⊥CA于H,BD=BE,CE=CF,求证:AD=AF.
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/15 11:16:55
已知:如图,设M是△ABC内部任意一点,MD⊥AB于G,ME⊥BC于K,MF⊥CA于H,BD=BE,CE=CF,求证:AD=AF.
已知:如图,设M是△ABC内部任意一点,MD⊥AB于G,ME⊥BC于K,MF⊥CA于H,BD=BE,CE=CF,求证:AD=AF.
已知:如图,设M是△ABC内部任意一点,MD⊥AB于G,ME⊥BC于K,MF⊥CA于H,BD=BE,CE=CF,求证:AD=AF.
连接MA、MB、MC
MD⊥AB,ME⊥BC,MF⊥CA
∴AF²=AH²+HF² (Rt△AHF)
=AM²-MH²+CF²-CH² (Rt△AMH:AH²=AM²-MH²,Rt△CFH:HF²=CF²-CH²
=AM²+CF²-(MH²+CH²)
=AM²+CF²-CM² (Rt△CMH:MH²+CH²=CM²)
=AM²+CE²-CM² (CF=CE)
AD²=DG²+AG² (Rt△ADG)
=BD²-BG²+AM²-MG² (Rt△BDG:DG²=BD²-BG²,Rt△AGM:AG²=AM²-MG²)
=AM²-(BG²+MG²)+BD²
=AM²-BM²+BE² (Rt△BMG:BG²+MG²=BM²,BD=BE)
=AM²-(BK²+MK²)+(BK²+EK²) (Rt△BMK:BM²=BK²+MK²,Rt△BEK:BE²=BK²+EK²)
=AM²-BK²-MK²+BK²+EK²
=AM²-MK²+EK²
=AM²-MK²+CE²-CK² (Rt△CEK:EK²=CE²-CK²)
=AM²+CE²-(MK²+CK²)
=AM²+CE²-CM² (Rt△CMK:MK²+CK²=CM²)
∴AF²=AD²
∴AF=AD
∵MD⊥AB于G,
∴AD^-BD^=MA^-MB^,
同理,BE^-CE^=MB^2-MC^,
CF^-AF^=MC^-MA^,
三式相加,把BD=BE,CE=CF得AD^-AF^=0,
∴AD=AF.