∠AOB的平分线OP,∠BOC的平分线OQ,则
∠POQ=∠POB+∠BOQ=(1/2)(∠AOB+∠BOC)
=(1/2)*180°=90°
∴PO⊥QO,即邻补角的平分线互相垂直三角形ABC中，D、E、F分别为BC、AC、AB的中点，AD、BE、CF相交于点0，AB=6，BC=10，AC=8.试求出线段DE、OA、OF的长度与
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