AB and CD are perpendicular to line BD with AB = 997,BD = 2012 and CD = 512.If E is a point on BD,what is the shortest possible distance forAE + EC?

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AB and CD are perpendicular to line BD with AB = 997,BD = 2012 and CD = 512.If E is a point on BD,what is the shortest possible distance forAE + EC?
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AB and CD are perpendicular to line BD with AB = 997,BD = 2012 and CD = 512.If E is a point on BD,what is the shortest possible distance forAE + EC?
AB and CD are perpendicular to line BD with AB = 997,BD = 2012 and CD = 512.If E is a point on BD,what is the shortest possible distance forAE + EC?

AB and CD are perpendicular to line BD with AB = 997,BD = 2012 and CD = 512.If E is a point on BD,what is the shortest possible distance forAE + EC?
作C点关于BD的对称点C',则有EC=EC', 连接AC' 交BD于点P
:. AE+EC=AE+EC'
根据三角形边长定理,当AEC'三点共线时,AE+EC'最短.

首先翻译一下吧，相信你看图也可以看懂
AB,CD分别垂直于BD与B,D。AB=997，BD=2012，CD=512.E为BD上一点，求AE+EC的最小值
这道题的做法如下
延长AB到F，使得AB=BF，连接FC与BD的交点即为所求点E的位置（利用对称性质及两点间线段最短）
此时AE+EC=FE+EC=FC
做CG垂直于AF于G，则三角形GFC中，FG=GB+...

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首先翻译一下吧，相信你看图也可以看懂
AB,CD分别垂直于BD与B,D。AB=997，BD=2012，CD=512.E为BD上一点，求AE+EC的最小值
这道题的做法如下
延长AB到F，使得AB=BF，连接FC与BD的交点即为所求点E的位置（利用对称性质及两点间线段最短）
此时AE+EC=FE+EC=FC
做CG垂直于AF于G，则三角形GFC中，FG=GB+BF=CD+AB=1509,CG=BD=2012
则此时由勾股定理可得FC=2515
希望对你有所帮助。

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