Question

如图，AD与BC相交于点O，OA等于OC，角A等于角C，BE等于DE.求证OE垂直平分BD.

Answer 1

∵∠A=∠C，OA=OC，∠AOB=∠COD ∴△ABO≌△CDO ∴OB=OD
∵OB=OD，BE=DE，OE=OE ∴△OBE≌△ODE ∴∠BOE=∠DOE
取OE于BD的交点为F
∵BO=DO，∠BOE=∠DOE，OF=OF ∴△BOF≌△DOF ∴BF=DF，∠OFB=∠OFD
∵∠OFB+∠OFD=180° ∴∠OFB=∠OFD=90° ∴OE垂直平分BD

Answer 2

∵∠A=∠C，OA=OC，∠AOB=∠COD ∴△ABO≌△CDO ∴OB=OD
∵OB=OD，BE=DE，OE=OE ∴△OBE≌△ODE ∴∠BOE=∠DOE
取OE于BD的交点为F
∵BO=DO，∠BOE=∠DOE，OF=OF ∴△BOF≌△DOF ∴BF=DF，∠OFB=∠OFD
∵∠OFB+∠OFD=180° ∴∠OFB=∠OFD=90° ∴OE垂直平分BD
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