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初中数学八年级下册(648题)


如图,在平行四边形\( ABCD\)中,\(AE \bot BC\)于点\(E\),延长\(BC\)\(F\)点使\(CF = BE\),连接\(DF\)。求证:四边形\(AEFD\)是矩形。

图片 12    



知识点:第十八章 平行四边形


参考答案:证明:
\(\because CF = BE\),
\(\therefore CF + EC = BE + EC\).即\(EF = BC\),
\(\because \)四边形\(ABCD\)是平行四边形,
\(\therefore AD//BC\),\(AD = BC\),
\(\therefore AD//EF\),\(AD = EF\),
\(\therefore \)四边形\(AEFD\)是平行四边形。
\(\because AE \bot BC\),
\(\therefore \angle AEF = 90^\circ \).
\(\therefore \)平行四边形\(AEFD\)是矩形。

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