如图,在梯形ABCD中,AD∥BC,AB=CD=BC=6,AD=3,点M为边BC的中点,以M为顶点作∠EMF=∠B射线ME交AB于点E 射线MF交腰CD于点F,连接EF 1 求证:△MEF∽△BEM2 若△BEM是以BM为腰的等腰三角形,求EF的长3 若EF⊥CD,求BE
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![如图,在梯形ABCD中,AD∥BC,AB=CD=BC=6,AD=3,点M为边BC的中点,以M为顶点作∠EMF=∠B射线ME交AB于点E 射线MF交腰CD于点F,连接EF 1 求证:△MEF∽△BEM2 若△BEM是以BM为腰的等腰三角形,求EF的长3 若EF⊥CD,求BE](/uploads/image/z/11230226-26-6.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E6%A2%AF%E5%BD%A2ABCD%E4%B8%AD%2CAD%E2%88%A5BC%2CAB%3DCD%3DBC%3D6%2CAD%3D3%2C%E7%82%B9M%E4%B8%BA%E8%BE%B9BC%E7%9A%84%E4%B8%AD%E7%82%B9%2C%E4%BB%A5M%E4%B8%BA%E9%A1%B6%E7%82%B9%E4%BD%9C%E2%88%A0EMF%3D%E2%88%A0B%E5%B0%84%E7%BA%BFME%E4%BA%A4AB%E4%BA%8E%E7%82%B9E+%E5%B0%84%E7%BA%BFMF%E4%BA%A4%E8%85%B0CD%E4%BA%8E%E7%82%B9F%2C%E8%BF%9E%E6%8E%A5EF+1+%E6%B1%82%E8%AF%81%EF%BC%9A%E2%96%B3MEF%E2%88%BD%E2%96%B3BEM2+%E8%8B%A5%E2%96%B3BEM%E6%98%AF%E4%BB%A5BM%E4%B8%BA%E8%85%B0%E7%9A%84%E7%AD%89%E8%85%B0%E4%B8%89%E8%A7%92%E5%BD%A2%2C%E6%B1%82EF%E7%9A%84%E9%95%BF3+%E8%8B%A5EF%E2%8A%A5CD%2C%E6%B1%82BE)
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