∵OA=OB∴∠OAB=∠OBA∵CD//AB∴∠OAB=∠C,∠OBA=∠D∴∠C=∠D∴OD=OC=OA+AC=3+2=5作OE⊥CD,交CD于E∵tan∠C =½,即OE/CE=½∴CE=2OE根据勾股定理：OC²=OE²+CE²5²=5OE²OE=√5连接O...