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

Question

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

Answer 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△ABC
OQ/CQ=S△OAB/S△ABC
∴OM/AM+ON/BN+OQ/CQ=(S△OBC+S△OAC+S△OAB)/S△ABC=S△ABC/S△ABC=1