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初中数学八年级上册试题库(555题)


第141题

如图,给出四个条件:\(AE\)平分\(\angle BAD\)\(BE\)平分\(\angle ABC\)\(AE \bot EB\)\(AB = AD + BC\).请你以其中3个作为条件,写出一个能推出\(AD//BC\)的正确命题,并给以证明。

图片 4



参考答案:见解析


解析:

解:

①②③\( \Rightarrow AD//BC\)

      \(\because \)\(AE \bot EB\)

      \(\therefore \angle EAB + \angle EBA = 90^\circ \)

\(\because \)\(AE\)平分\(\angle BAD\)\(BE\)平分\(\angle ABC\)

      \(\therefore \angle EAB = \frac{1}{2}\angle DAB\)\(\angle EBA = \frac{1}{2}\angle CBA\)

      \(\therefore \angle DAB + \angle CBA = 180^\circ \)

      \(\therefore AD//BC\)

图片 2

①②④\( \Rightarrow AD//BC\)

  \(AB\)上取点\(M\),使\(AM = AD\),连接\(EM\)

\(\because AB = AD + BC\)

\(\therefore MB = BC\)

\(\Delta AEM\)\(\Delta AED\)

\(\begin{cases}
AM = AD \\\
\angle MAE = \angle DAE \\\
AE = AE \\\
\end{cases}\)

\(\therefore \Delta AEM \cong \Delta AED(SAS)\)

\(\therefore \angle D = \angle AME\)

\(\Delta BEM\)\(\Delta BEC\)

\(\begin{cases}
BM = BC \\\
\angle MBE = \angle CBE \\\
BE = BE \\\
\end{cases}\)

\(\therefore \Delta BEM \cong \Delta BEC(SAS)\)

\(\therefore \angle C = \angle BME\)

\(\therefore \)\(\angle D + \angle C = \angle AME + \angle BME = 180^\circ \)

\(\therefore AD//BC\)

①③④②③④都不能推出\(AD//BC\)













第153题

如图,有\(A\)\(B\)\(C\)三个居民小区的位置成三角形,现决定在三个小区之间修建一个购物超市,使超市到三个小区的距离相等,则超市应建在\((\)  \()\)


A.在\(AC\),\(BC\)两边高线的交点处

B.在\(AC\),\(BC\)两边中线的交点处

C.在\(AC\),\(BC\)两边垂直平分线的交点处

D.在\(\angle A\),\(\angle B\)两内角平分线的交点处在


参考答案:C









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