图中字母和条件中有些不一致，按照已知作图证明如下取BC的中点P，连接PM，PN∵P是BC的中点，M是AB的中点∴PM‖AC,PM=1/2AC∵N是CD的中点，∴PN=1/2BD,PN‖BD∵BD =AC ∴PM =PN∴∠PMN=∠PNM∵PM‖BD,PN‖AC∴∠PMN=∠OFE,∠PNM=∠OEF∴∠OEF=∠OFE∴OE=OF
