如图,已知AB=AC,∠A=36°,AB的中垂线MN交AC于点D,交AB于点M,有下面4个结论:①BD是∠ABC的角平分线; ②△BCD是等腰三角形; ③△ABC∽△BCD; ④△AMD≌△BCD.从你认为正确的结论中选出一个加以
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![如图,已知AB=AC,∠A=36°,AB的中垂线MN交AC于点D,交AB于点M,有下面4个结论:①BD是∠ABC的角平分线; ②△BCD是等腰三角形; ③△ABC∽△BCD; ④△AMD≌△BCD.从你认为正确的结论中选出一个加以](/uploads/image/z/8769871-55-1.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%B7%B2%E7%9F%A5AB%3DAC%2C%E2%88%A0A%3D36%C2%B0%2CAB%E7%9A%84%E4%B8%AD%E5%9E%82%E7%BA%BFMN%E4%BA%A4AC%E4%BA%8E%E7%82%B9D%2C%E4%BA%A4AB%E4%BA%8E%E7%82%B9M%2C%E6%9C%89%E4%B8%8B%E9%9D%A24%E4%B8%AA%E7%BB%93%E8%AE%BA%EF%BC%9A%E2%91%A0BD%E6%98%AF%E2%88%A0ABC%E7%9A%84%E8%A7%92%E5%B9%B3%E5%88%86%E7%BA%BF%EF%BC%9B+%E2%91%A1%E2%96%B3BCD%E6%98%AF%E7%AD%89%E8%85%B0%E4%B8%89%E8%A7%92%E5%BD%A2%EF%BC%9B+%E2%91%A2%E2%96%B3ABC%E2%88%BD%E2%96%B3BCD%EF%BC%9B+%E2%91%A3%E2%96%B3AMD%E2%89%8C%E2%96%B3BCD%EF%BC%8E%E4%BB%8E%E4%BD%A0%E8%AE%A4%E4%B8%BA%E6%AD%A3%E7%A1%AE%E7%9A%84%E7%BB%93%E8%AE%BA%E4%B8%AD%E9%80%89%E5%87%BA%E4%B8%80%E4%B8%AA%E5%8A%A0%E4%BB%A5)
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