“微信扫一扫”进入考试题库练习及模拟考试
初中数学八年级上册试题库(555题)
A.1
B.1.5
C.2
D.2.5
参考答案：A
第83题
已知，
A.\(△ABC≌△XYZ\)
B.\(△DEF≌△XYZ\)
C.\(∠C＝∠Z\)
D.\(∠F＝80°\)
参考答案：C
参考答案：
证明：连接
在
{AC = BD} \\
{BC = CB}
\end{array}} \right.\)
在
\begin{gathered}
\angle A = \angle D \\
\angle AEB = \angle DEC \\
\end{gathered} \\
{AB = DC}
\end{array}} \right.\)
声明：试题解析著作权属菁优网所有，未经书面同意，不得复制发布日期：2022/7/17 15:34:21；用户：邓芸；邮箱：syxx019@xyh.com；学号：21059015
A.\(\Delta ABC \cong \Delta CDE\)
B.\(\angle CAB = \angle DCE\)
C.\(AB \bot CD\)
D.\(E\)为\(BC\)中点
参考答案：D
A.\(49^\circ \)
B.\(59^\circ \)
C.\(41^\circ \)
D.\(51^\circ \)
参考答案：C
A.\(50^\circ \)
B.\(55^\circ \)
C.\(60^\circ \)
D.\(70^\circ \)
参考答案：C
A.6
B.7
C.8
D.9
参考答案：C
第89题
如图，在
A.5
B.6
C.7
D.8
参考答案：A
A.1
B.2
C.3
D.5
参考答案：B
第91题
如图，在
A.\(\angle 1 = \angle 2\)
B.\(\angle 2 = 2\angle 1\)
C.\(\angle 1 + \angle 2 = 90^\circ \)
D.\(\angle 1 + \angle 2 = 180^\circ \)
参考答案：D
第92题
如图，点
A.6
B.7
C.8
D.9
参考答案：A
参考答案：见解析
解析：
证明：延长
在
在
参考答案：证明：\(\because \angle B = \angle DCF\)，\(\therefore CF//AB\)，\(\therefore \angle FCA = \angle A\)，\(\angle F = \angle FGA\)，\(\because \)点\(E\)是\(AC\)的中点，\(\therefore AE = EC\)，在\(\Delta AEG\)和\(\Delta CEF\)中，\(\left\{ {\begin{array}{*{20}{l}}
{\angle A = \angle FCA} \\
{\angle FGA = \angle F} \\
{AE = EC}
\end{array}} \right.\)，\(\therefore \Delta AEG \cong \Delta CEF(AAS)\)。\(\because \Delta AEG \cong \Delta CEF\)；\(\therefore CF = AG = 6\)，\(\because AG = 3BG\)，\(\therefore BG = 2\)，\(\therefore AB = 8\)，\(\therefore \Delta ABC\)的周长\( = AB + AC + BC = 28\)。
参考答案：线段\(AD\)与\(AG\)垂直且相等\(\because BE \bot AC\)，\(CF \bot AB\)，\(\therefore \angle HFB = \angle HEC = 90^\circ \)，又\(\because \angle BHF = \angle CHE\)，\(\therefore \angle ABD = \angle ACG\)，在\(\Delta ABD\)和\(\Delta GCA\)中\(\left\{ {\begin{array}{*{20}{l}}
{AB = CG} \\
{\angle ABD = \angle ACG} \\
{BD = CA}
\end{array}} \right.\)，\(\therefore \Delta ABD \cong \Delta GCA(SAS)\)，\(\therefore AD = GA\)（全等三角形的对应边相等）；\(\because \Delta ABD \cong \Delta GCA\)，\(\therefore \angle ADB = \angle GAC\)，又\(\because \angle ADB = \angle AED + \angle DAE\)，\(\angle GAC = \angle GAD + \angle DAE\)，\(\therefore \angle AED = \angle GAD = 90^\circ \)，\(\therefore AD \bot GA\)。