《2025年思维新观察八年级数学上册人教版湖北专版》

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:$\because \angle AOB = \angle COD$,
$\therefore \angle AOB + \angle AOC = \angle COD + \angle AOC$,即$\angle BOC = \angle AOD$,
在$\triangle AOD$和$\triangle BOC$中,$\left\{ \begin{array}{l} OA = OB \\ \angle BOC = \angle AOD, \\ OC = OD \end{array} \right.$
$\therefore \triangle AOD \cong \triangle BOC$(SAS);
(2)求$\angle AEB$的度数.
解:由(1)知$\triangle AOD \cong \triangle BOC$,
$\therefore \angle A = \angle B$,
设$OA$与$BC$交于点$F$,$\because \angle AFE = \angle BFO$,
$\therefore \angle AEB = \angle AOB =$$90^{\circ}$.

答案： 证明:
(1) $\because \angle AOB = \angle COD$,
$\therefore \angle AOB + \angle AOC = \angle COD + \angle AOC$,
$\left\{ \begin{array}{l} OA = OB \\ \angle BOC = \angle AOD, \\ OC = OD \end{array} \right.$
$\therefore \triangle AOD \cong \triangle BOC$;
(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,
$\left\{ \begin{array}{l} AC = DC \\ \angle ACE = \angle DCB, \\ CE = CB \end{array} \right.$
$\therefore \triangle ACE \cong \triangle DCB$,
$\therefore \angle CAE = \angle CDB$,
$\therefore \angle DFA = \angle ACD = 60^{\circ}$, $\therefore \angle AFB = 120^{\circ}$;
(2) 证$\triangle ACE \cong \triangle DCB$,
$\angle AEC = \angle DBC$,
则$\angle EFB = \angle ECB = \alpha$,
$\therefore \angle AFB = 180^{\circ} - \alpha$.