在△ABC中,AB=AC,∠A=120°,∠AB的垂直平分线MN分别交BC AB与点M N,求证:CM=2BC
答得好,
人气:210 ℃ 时间:2019-11-05 05:10:37
解答
题目有问题吧,既然M是BC上一点,明显BC>CM
应该是CM=2BM吧
因为AB=AC,∠A=120度
所以∠B=∠C=30度
连接AM,因MN是AB的垂直平分线
所以AM=BM,∠BAM=∠B=30度
所以∠CAM=120-30=90度
所以在直角三角形AMC中,CM=2AM
即CM=2BM
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