AH是三角形ABC的高,D、E、F分别是AB、BC、CA的中点,延长DF到G,使FG=EH,求证 AH和DG 互相垂直平分

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AH与DG交于点O
容易证明DG垂直于AH且平分AH，所以要想证明结论，只需证明OD=OG即可
OD= BH/2 = (BE + EH)/2 = (BC/2 + EH)/2 = BC/4 + EH/2
OG= OF + FG = OF + EH = CH/2 + EH = (CE - EH)/2 + EH = (BC/2 - EH)/2 + EH = BC/4 + EH/2
所以OD = OG
所以AH和DG互相垂直平分

由DF是 ΔABC的中位线,所以DF=1/2BC=BE,
又FG=EH,所以DG=BH.而DG‖BH
所以DBHG是平行四边形
所以HG‖BD,HG=BD即HG‖AD,HG=AD
所以ADHG是平行四边形,AD和DG互相平分
易证AH⊥BC而DG‖BC
所以AH⊥DG
所以AH和DG互相垂直平分.