请解析,如图,在五边形ABCDE中,∠BAE=120°...如下如图,在五边形ABCDE中,∠BAE=120°,∠B=∠E=90°.AB=BC,AE=DE,在BC,DE上分别找一点M,N,使得△AMN的周长最小时,则∠AMN+∠ANM的度数为( )A.100° B.110° C.120° D.1
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![请解析,如图,在五边形ABCDE中,∠BAE=120°...如下如图,在五边形ABCDE中,∠BAE=120°,∠B=∠E=90°.AB=BC,AE=DE,在BC,DE上分别找一点M,N,使得△AMN的周长最小时,则∠AMN+∠ANM的度数为( )A.100° B.110° C.120° D.1](/uploads/image/z/2122366-22-6.jpg?t=%E8%AF%B7%E8%A7%A3%E6%9E%90%2C%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E4%BA%94%E8%BE%B9%E5%BD%A2ABCDE%E4%B8%AD%2C%E2%88%A0BAE%3D120%C2%B0...%E5%A6%82%E4%B8%8B%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E4%BA%94%E8%BE%B9%E5%BD%A2ABCDE%E4%B8%AD%2C%E2%88%A0BAE%3D120%C2%B0%2C%E2%88%A0B%3D%E2%88%A0E%3D90%C2%B0.AB%3DBC%2CAE%3DDE%2C%E5%9C%A8BC%2CDE%E4%B8%8A%E5%88%86%E5%88%AB%E6%89%BE%E4%B8%80%E7%82%B9M%2CN%2C%E4%BD%BF%E5%BE%97%E2%96%B3AMN%E7%9A%84%E5%91%A8%E9%95%BF%E6%9C%80%E5%B0%8F%E6%97%B6%2C%E5%88%99%E2%88%A0AMN%2B%E2%88%A0ANM%E7%9A%84%E5%BA%A6%E6%95%B0%E4%B8%BA%EF%BC%88+%EF%BC%89A.100%C2%B0+B.110%C2%B0+C.120%C2%B0+D.1)
请解析,如图,在五边形ABCDE中,∠BAE=120°...如下如图,在五边形ABCDE中,∠BAE=120°,∠B=∠E=90°.AB=BC,AE=DE,在BC,DE上分别找一点M,N,使得△AMN的周长最小时,则∠AMN+∠ANM的度数为( )A.100° B.110° C.120° D.1
请解析,如图,在五边形ABCDE中,∠BAE=120°...如下
如图,在五边形ABCDE中,∠BAE=120°,∠B=∠E=90°.
AB=BC,AE=DE,在BC,DE上分别找一点M,N,使得△AMN的周长最小时,则∠AMN+∠ANM的度数为( )
A.100° B.110° C.120° D.130°
请解析,如图,在五边形ABCDE中,∠BAE=120°...如下如图,在五边形ABCDE中,∠BAE=120°,∠B=∠E=90°.AB=BC,AE=DE,在BC,DE上分别找一点M,N,使得△AMN的周长最小时,则∠AMN+∠ANM的度数为( )A.100° B.110° C.120° D.1
∠AMN+∠ANM=120°
延长AB到A'使BA'=AB,
延长AE到A''使AE=EA'',
连接A'M,A''N
△AA‘M中;AB=BA’;MB⊥AA';
因此MB是垂直平分线;故此:
AM=A'M,同理A''N=AN
折线A'M,NM,A''N即为△AMN的周长
根据两点之间直线最短,M,N点在直线A'A''上
此时有
角MA’A=∠MAA‘;
同理可得:NE是AA’‘的垂直平分线;
∠NAA''=∠NA’‘A;
而∠A’AA‘’=120°;
所以∠AA‘A’‘=∠AA’‘A=30°;
所求的两个角:∠AMN+∠ANM=2∠A'+2∠A''=2(180-∠BAE)=120°