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[几何] 梯形中证明等腰直角三角形

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乌贼 Posted 2015-6-1 21:35 |Read mode
转自   lanqi.org/?p=1201     ( 证求純几)
   如图:梯形$ ABCD $中,$ AD\px BC $,$ AC,BD $交于$ E $,$ BC=BD,CD=CE,\angle ABD=15\du  $。求证:$ \triangle ABC $为等腰直角三角形。
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isee Posted 2015-6-2 21:06
同一法,倒是简单。
先作等腰直角三角形ABC,过A作BC的平行线,在其平行上取点D,使BD=BC.
然后证明角DBC为30度,是此图符合原题。

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其妙 Posted 2015-6-2 22:54
回复 2# isee
陈题吧?。

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 Author| 乌贼 Posted 2015-6-2 23:40
回复 3# 其妙
给个链接

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其妙 Posted 2015-6-3 19:33
我说的IC的方法涉及到的题

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