《2025年诚成教育学业评价八年级数学下册人教版》

25.(8分)如图，在△ABC中，AB = 3，AC = 5，边BC上的中线AD = 2，延长AD到点E，使DE = AD，连接CE.
(1)求证AE⊥CE;
(2)求BC边的长.
　　

答案：
(1)证明：$\because D$为$BC$的中点，$\therefore BD = CD$.
又$\angle ADB=\angle EDC$，$AD = ED$
$\therefore \triangle ABD\cong\triangle ECD$
$\therefore CE = AB = 3$.
$\because AE = 2AD = 4$，$AC = 5$
$\therefore AC^{2}=AE^{2}+CE^{2}$
$\therefore \angle E = 90^{\circ}$
$\therefore AE\perp CE$.
(2)解：在$Rt\triangle CDE$中，由勾股定理，得$CD=\sqrt{CE^{2}+DE^{2}}=\sqrt{3^{2}+2^{2}}=\sqrt{13}$
$\therefore BC = 2CD = 2\sqrt{13}$.