证明：过点A作AE⊥BC,过点D作DF⊥BC
∵AE⊥BC,DF⊥BC,AD∥BC
∴AE＝DF
∵S△ABC＝BC×AE/2,S△DBC＝BC×DF/2
∴S△ABC＝S△DBC
∵S△ABC＝S△AOB+ S△BOC,S△DBC＝S△COD+ S△BOC
∴S△AOB＝S△COD