设线段AB长度为x,则 x + 2/3x = 10解得x = 6
所以线段AB = 6, BD = 4, DE = 4, EC = 4. AE = 14
将线段AB延长到点C,使BC=2AB,画线段BC的三等分点D,E(BD小于BE),已知AD=10cm,求AB,AE的长!
将线段AB延长到点C,使BC=2AB,画线段BC的三等分点D,E(BD小于BE),已知AD=10cm,求AB,AE的长!
数学人气:551 ℃时间:2020-04-11 13:11:15
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