《2025年思维新观察八年级数学上册人教版》

9. 如图,$\angle ACD= \angle BCE= 20^{\circ}$,$CA= CD$,$CB= CE$,$AB与DE相交于点M$,则$\angle BME$的度数为______20°.

答案： $20^{\circ}$

10. 如图,$\triangle ABC$,$\angle A= \angle B$,点$D在AB$上,$\angle EDF= \alpha$,交$AC于点E$,$BC于点F$,$AD= BF$,$AE= BD$,则下列结论正确的有______①②③(填序号).
①$\triangle ADE\cong\triangle BFD$； ②$\angle BDF= \angle AED$；③$\angle C= 180^{\circ}-2\alpha$； ④$\angle ADE= \angle B$.

答案： ①②③

11. 如图,$OA= OB$,$OC= OD$,$\angle AOB= \angle COD= 90^{\circ}$,$AD$,$BC交于点E$.
(1) 求证：$\triangle AOD\cong\triangle BOC$；
证明:$\angle COB = \angle AOD$,$\triangle AOD \cong \triangle BOC(SAS)$
(2) 求$\angle AEB$的度数.
$90^{\circ}$

答案： 证明:
(1)$\angle COB = \angle AOD$,$\triangle AOD \cong \triangle BOC(SAS)$;
(2)$\triangle AOD \cong \triangle BOC$,
$\therefore \angle A = \angle B$,
$\therefore \angle AEB = \angle AOB = 90^{\circ}$.

12. 利用无刻度尺作图：$A$,$B$,$C$为格点,在下列图中找格点$D$,连$CD$,使$CD\perp AB$.

答案：
如图所示

13. 已知点C为线段AB上一点,分别以AC,BC为边在线段AB的同侧作$\triangle ACD$和$\triangle BCE,$且$CA= CD,CB= CE,\angle ACD= \angle BCE,$直线AE与BD交于点F.(1) 如图1,若$\angle ACD= 60^{\circ},$求$\angle AFB$的度数；120°(2) 如图2,将图1中的$\triangle ACD$绕点C顺时针旋转(交点F在BD的延长线上),若$\angle ACD= \alpha,$试探究$\angle AFB$与$\alpha$的数量关系,并予以证明.$\angle AFB = 180^{\circ} - \alpha$

答案： 解:
(1)$\because \angle ACD = \angle BCE$,$\therefore \angle ACE = \angle BCD$,
在$\triangle ACE$和$\triangle DCB$中,
$\left\{ \begin{array}{l} AC = DC \\ \angle ACE = \angle DCB \\ CE = CB \end{array} \right. $
$\therefore \triangle ACE \cong \triangle DCB$,
$\therefore \angle CAE = \angle CDB$,
$\therefore \angle DFA = \angle ACD = 60^{\circ}$,$\angle AFB = 120^{\circ}$;
(2)证$\triangle ACE \cong \triangle DCB$,
$\angle AEC = \angle DBC$,
则$\angle EFB = \angle ECB = \alpha$,
$\therefore \angle AFB = 180^{\circ} - \alpha$.