《2025年大联考单元期末测试卷八年级数学上册沪科版》

21. 如图,在$ \triangle ABC $中,$ BA = BC $,点$ D $在边$ CB $上,且$ DB = DA = AC $.
(1)如图1,填空:$ \angle B = $$36^{\circ}$,$ \angle C = $$72^{\circ}$.
(2)若点$ M $为线段$ BC $上的点,过点$ M $作直线$ MH \perp AD $于点$ H $,分别交直线$ AB $,$ AC $于点$ N $,$ E $,如图2.
①求证:$ \triangle ANE $是等腰三角形;
②试写出线段$ BN $,$ CE $,$ CD $之间的数量关系,并证明.

答案： 21.解：
(1)$\because BA = BC$，$\therefore\angle BCA = \angle BAC$。
$\because DA = DB$，$\therefore\angle BAD = \angle B$。
$\because AD = AC$，$\therefore\angle ADC = \angle C = \angle BAC = 2\angle B$。(2分)
$\because\angle B + \angle BAC + \angle C = 180^{\circ}$，$\therefore\angle B + 2\angle B + 2\angle B = 180^{\circ}$，
$\therefore\angle B = 36^{\circ}$，$\angle C = 2\angle B = 72^{\circ}$。(4分)
(2)①证明：在$\triangle ADB$中，$\because DB = DA$，$\angle B = 36^{\circ}$，$\therefore\angle BAD = 36^{\circ}$。
$\because\angle BAC = 2\angle B$，$\therefore\angle BAC = 72^{\circ}$，$\angle CAD = 36^{\circ}$，$\therefore\angle BAD = \angle CAD$。
$\because MH\perp AD$，$\therefore\angle AHN = \angle AHE = 90^{\circ}$。(7分)
在$\triangle ANH$和$\triangle AEH$中，$\because\begin{cases}\angle BAD = \angle CAD,\\AH = AH,\\\angle AHN = \angle AHE,\end{cases}$
$\therefore \triangle ANH\cong \triangle AEH(ASA)$，$\therefore AN = AE$，即$\triangle ANE$是等腰三角形。(9分)
②$CD = BN + CE$。
证明：由①知$AN = AE$。
又$\because BA = BC$，$DB = AC$，$\therefore BN = AB - AN = BC - AE$，$CE = AE - AC = AE - DB$，
$\therefore BN + CE = BC - DB = CD$，即$CD = BN + CE$。(12分)