三角形ABC中任意一点O 作OA,OB,OC的延长经交BC,AC,AB于点M,N,Q,证明OM/AM+ON/BN+OQ/CQ=1

作AD⊥BC,OH⊥BC.则⊿OMH∽⊿AMD∴OH/AD=OM/AM ∵OH/AD===⊿OBC/⊿ABC∴OM/AM==⊿OBC/⊿ABC同理,ON/BN=S△OAC/S△ABCOQ/CQ=S△OAB/S△ABC∴OM/AM+ON/BN+OQ/CQ=(S△OBC+S△OAC+S△OAB)/S△ABC=S△ABC/S△ABC=1...