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初中数学八年级下册(648题)


如图,在\(\Delta ABC\)中,\(AB = AC\),点\(D\)\(E\)分别是线段\(BC\)\(AD\)的中点,过点\(A\)\(BC\)的平行线交\(BE\)的延长线于点\(F\),连接\(CF\)。求证:四边形\(ADCF\)是矩形。

图片 1



知识点:第十八章 平行四边形


参考答案:证明:\(\because AF//BC\)

\(\therefore \angle AFE = \angle DBE\)

\(\because E\)是线段\(AD\)的中点

\(\therefore AE = DE\)

\(\because \angle AEF = \angle DEB\)

\(\therefore \Delta BDE\cong \Delta FAE(\text{AAS})\)

\(\therefore AF = BD\)

\(\because D\)是线段\(BC\)的中点

\(\therefore BD = CD\)

\(\therefore AF = CD\)

\(\because AF//CD\)

\(\therefore \)四边形\(ADCF\)是平行四边形

\(\because AB = AC\)

\(\therefore AD \bot BC\)

\(\therefore \angle ADC = 90^\circ \)

\(\therefore \)平行四边形\(ADCF\)为矩形

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