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:A

2个回答

  • 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