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
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