如图,点E在正方形ABCD对角线BD所在的直线上,过点E作EF⊥AE,交直线BC于点F,EG⊥BC,垂足为点G.(1)第一问也证明一下吧.
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![如图,点E在正方形ABCD对角线BD所在的直线上,过点E作EF⊥AE,交直线BC于点F,EG⊥BC,垂足为点G.(1)第一问也证明一下吧.](/uploads/image/z/3186424-64-4.jpg?t=%E5%A6%82%E5%9B%BE%2C%E7%82%B9E%E5%9C%A8%E6%AD%A3%E6%96%B9%E5%BD%A2ABCD%E5%AF%B9%E8%A7%92%E7%BA%BFBD%E6%89%80%E5%9C%A8%E7%9A%84%E7%9B%B4%E7%BA%BF%E4%B8%8A%2C%E8%BF%87%E7%82%B9E%E4%BD%9CEF%E2%8A%A5AE%2C%E4%BA%A4%E7%9B%B4%E7%BA%BFBC%E4%BA%8E%E7%82%B9F%2CEG%E2%8A%A5BC%2C%E5%9E%82%E8%B6%B3%E4%B8%BA%E7%82%B9G.%EF%BC%881%EF%BC%89%E7%AC%AC%E4%B8%80%E9%97%AE%E4%B9%9F%E8%AF%81%E6%98%8E%E4%B8%80%E4%B8%8B%E5%90%A7.)
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