四边形ABDC中,AB=AC,BD=BC,平面内两动点点E,F满足∠EAF=1/2∠BAC,∠EDF=1/2∠BDC,BE, CF交于G,求证:
(1)S△ABE+S△ACF+S△DEF=S△DBE+S△DCF+S△AEF;
(2)AG/DG为一定值.
证明:(1)将△ABE绕A旋转得△ACM,将△DBE绕D旋转得△DCN.
则∠EAF=∠MAF,∠EDF=∠NDF,
△EAF≌△MAF,△EDF≌△HDF,
则FM=FE=FN,于是△FCM≌△FCN,从而S△ABE+S△ACF+S△DEF=S△DBE+S△DCF+S△AEF,得证.
(2)将△ABG绕A旋转得△ACM',将△DBG绕D旋转得△DCN'.
易得△GCM'≌△GCN',GM'=GN'
又易得△ABC∽△AGM,△DBC∽△DGN,
所以AB/BC=AG/GM',DB/BC=DG/GN',
故AG/DG=AB/DB,为一定值.
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