1、已知,如图正方形ABCD,E是BC的重点,CF为∠BCD的外角∠GCD的平分线,EF⊥AE.求证:AE=EF.2、正方形ABCD,E为AC上的任一点,现作EF⊥AB於F,作EG⊥BC於G,试猜测DE与FG的关系,并说明理由【图传不上来,请到
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![1、已知,如图正方形ABCD,E是BC的重点,CF为∠BCD的外角∠GCD的平分线,EF⊥AE.求证:AE=EF.2、正方形ABCD,E为AC上的任一点,现作EF⊥AB於F,作EG⊥BC於G,试猜测DE与FG的关系,并说明理由【图传不上来,请到](/uploads/image/z/10908671-23-1.jpg?t=1%E3%80%81%E5%B7%B2%E7%9F%A5%2C%E5%A6%82%E5%9B%BE%E6%AD%A3%E6%96%B9%E5%BD%A2ABCD%2CE%E6%98%AFBC%E7%9A%84%E9%87%8D%E7%82%B9%2CCF%E4%B8%BA%E2%88%A0BCD%E7%9A%84%E5%A4%96%E8%A7%92%E2%88%A0GCD%E7%9A%84%E5%B9%B3%E5%88%86%E7%BA%BF%2CEF%E2%8A%A5AE.%E6%B1%82%E8%AF%81%EF%BC%9AAE%3DEF.2%E3%80%81%E6%AD%A3%E6%96%B9%E5%BD%A2ABCD%2CE%E4%B8%BAAC%E4%B8%8A%E7%9A%84%E4%BB%BB%E4%B8%80%E7%82%B9%2C%E7%8E%B0%E4%BD%9CEF%E2%8A%A5AB%E6%96%BCF%2C%E4%BD%9CEG%E2%8A%A5BC%E6%96%BCG%2C%E8%AF%95%E7%8C%9C%E6%B5%8BDE%E4%B8%8EFG%E7%9A%84%E5%85%B3%E7%B3%BB%2C%E5%B9%B6%E8%AF%B4%E6%98%8E%E7%90%86%E7%94%B1%E3%80%90%E5%9B%BE%E4%BC%A0%E4%B8%8D%E4%B8%8A%E6%9D%A5%2C%E8%AF%B7%E5%88%B0)
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