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题目漏掉了条件"AB垂直CD",若不加上则无法证出结论.证明:连接DG.CD为直径,则∠CGD=90°.∵∠COE=∠CGD=90°;∠OCE=∠GCD.∴⊿COE∽⊿CGD,GD/GC=OE/OC;又OE/OC=OE/OB=1/2,则GD/GC=1/2.∵∠CGA=(1/2)∠COA=45°.∴∠CGA=∠DGA=45°,DH/CH=GD/GC=1/2.(三角形内角平分线定理)则DH/DC=1/3,DH=DC/3; OH=OD-DH=DC/2-DC/3=DC/6.所以,OH:OD=(DC/6):DC/2)=1:3,即点O为OD的三等分点.... |
