已知:在正方形ABCD中,AC,BD交于点O,延长CB到点E,使BE=CB,连接DE交AB于F,求证:OF=二分之一BE.图:
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![已知:在正方形ABCD中,AC,BD交于点O,延长CB到点E,使BE=CB,连接DE交AB于F,求证:OF=二分之一BE.图:](/uploads/image/z/5094918-54-8.jpg?t=%E5%B7%B2%E7%9F%A5%EF%BC%9A%E5%9C%A8%E6%AD%A3%E6%96%B9%E5%BD%A2ABCD%E4%B8%AD%2CAC%2CBD%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%3DCB%2C%E8%BF%9E%E6%8E%A5DE%E4%BA%A4AB%E4%BA%8EF%2C%E6%B1%82%E8%AF%81%EF%BC%9AOF%3D%E4%BA%8C%E5%88%86%E4%B9%8B%E4%B8%80BE.%E5%9B%BE%EF%BC%9A)
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