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2012-05-04 15:53:52| 分类：试题集锦 | 标签： |举报 |字号大中小订阅

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本题为2012年《学在荆州》中考数学复习资料的P55第10题。下面的解答比原答案要简洁一些，故贴出同学们阅读。

如图，分别以Rt△ABC的直角边AC及斜边AB向外作等边△ACD、等边△ABE.已知∠BAC＝30°，EF⊥AB，垂足为F，连接DF.

(1)试说明AC＝EF；

(2)求证：四边形ADFE是平行四边形．

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解　(1)在Rt△ABC中，∠BAC＝30°，

∴BC＝AB，AC＝AB.

在等边△ABE中，EF⊥AB，

∴∠AFE＝90°，AF＝AE，EF＝AE＝AB，

∴AC＝EF.

(2)在等边△ACD中，∠DAC＝60°，

∴∠DAF＝60°＋30°＝90°＝∠EFA，

∴AD∥EF.

又AD＝AC＝EF，

∴四边形ADEF是平行四边形．

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