令AD的中点为G.由三角形中位线定理,有:
FG∥BD,FG=BD/2, EG∥CA,EG=AC/2.
∵OM=ON,∴∠OMN=∠ONM.
∵FG∥BD,∴∠GFE=∠OMN,∵EG∥CA,∴∠GEF=∠ONM.
由∠OMN=∠ONM,∠GFE=∠OMN,∠GEF=∠ONM,得:∠GFE=∠GEF,∴FG=EG.
由FG=BD/2,EG=AC/2,FG=EG,得:BD=AC.
令AD的中点为G.由三角形中位线定理,有:
FG∥BD,FG=BD/2, EG∥CA,EG=AC/2.
∵OM=ON,∴∠OMN=∠ONM.
∵FG∥BD,∴∠GFE=∠OMN,∵EG∥CA,∴∠GEF=∠ONM.
由∠OMN=∠ONM,∠GFE=∠OMN,∠GEF=∠ONM,得:∠GFE=∠GEF,∴FG=EG.
由FG=BD/2,EG=AC/2,FG=EG,得:BD=AC.