Question

已知正方形ABCD的对角线AC、BD相交于点O,E是OB上一点,DG⊥CE,垂足为G,DG交OC于点F,求EBCF是等腰

Answer 1

∵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.