正方形ABCD的对角线AC、BD相交于O点,E是BD上一点,DG⊥CE,垂足为G.DG交OC于F点.求证：四边形EBCF是等腰梯
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∵ABCD是正方形,∴∠COE＝90°,而DG⊥CE,∴∠EDG＝90°,得：∠COE＝∠EDG,
又∠CEO＝∠DEG,∴△CEO∽△DEG,得：∠ECO＝∠FDO.
∵ABCD是正方形,∴CO＝DO,∠COE＝∠DOF＝90°,而∠ECO＝∠FDO,
∴△COE≌△DOF,∴OE＝OF.
∵ABCD是正方形,∴OB＝OC,结合OE＝OF,得：OE/OB＝OF/OC,∴EF∥BC.
再由OB＝OC,OE＝OF,得：OB－OE＝OC－OF,∴BE＝CF.
由EF∥BC,BE＝CF,得：EBCF是等腰梯形.