如图,在正方形ABCD中,AC、BD交于点O,延长CB到点E,使BE=BC,连接DE交AB于点F
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![如图,在正方形ABCD中,AC、BD交于点O,延长CB到点E,使BE=BC,连接DE交AB于点F](/uploads/image/z/7850262-30-2.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E6%AD%A3%E6%96%B9%E5%BD%A2ABCD%E4%B8%AD%2CAC%E3%80%81BD%E4%BA%A4%E4%BA%8E%E7%82%B9O%2C%E5%BB%B6%E9%95%BFCB%E5%88%B0%E7%82%B9E%2C%E4%BD%BFBE%3DBC%2C%E8%BF%9E%E6%8E%A5DE%E4%BA%A4AB%E4%BA%8E%E7%82%B9F)
如图,在正方形ABCD中,AC、BD交于点O,延长CB到点E,使BE=BC,连接DE交AB于点F
如图,在正方形ABCD中,AC、BD交于点O,延长CB到点E,使BE=BC,连接DE交AB于点F
如图,在正方形ABCD中,AC、BD交于点O,延长CB到点E,使BE=BC,连接DE交AB于点F
BE=BC AD=BC
AD=BE
AD‖BC
∠E=∠ADE ∠ABE=∠BAD
⊿BFE≌⊿AFD
DF=EF
OB=OD
OF‖BE OF=1/2 BE
证明:∵四边形ABCD是正方形,
∴BC=AD.
又BE=BC,
∴BE=AD.
∵AD∥BE,
∴∠E=∠ADF,∠AFD=∠EFB.
∴△ADF≌△BEF.
∴DF=FE.
又∵DO=OB.
∴OF为△BDE的中位线.
∴OF=1 /2 BE.