《2025年启东中学作业本八年级数学上册苏科版连淮专版》

9.(2024·东海县期中)如图,在△ABC中,AC和BC的垂直平分线$l_1和l_2$分别交AB于点D,E.若AD= 3,DE= 4,EB= 5,则△ABC的面积等于______18.

答案： 18

10.如图,F是正五边形ABCDE中边DE的中点,连接BF并延长与CD的延长线交于点G,则∠BGC的度数为18°.

答案： $18 ^ { \circ }$

11.(15分)(2024·陕西)如图,已知直线l和l外一点A,请用尺规作图法,求作一个等腰直角△ABC,使得顶点B和顶点C都在直线l上.(作出符合题意的一个等腰直角三角形即可,保留作图痕迹,不写作法)

答案：
解:如答图所示,等腰直角$\triangle ABC$即为所求.

12.(15分)(2024·灌南县期中)如图,在△ABC中,AB= AC,D为BC边上一点,∠B= 30°,∠DAB= 45°.
(1)求∠DAC的度数；
(2)求证：DC= AB.

答案：
(1)解:$\because AB = AC$,$\therefore \angle C = \angle B = 30 ^ { \circ }$.
$\because \angle C + \angle BAC + \angle B = 180 ^ { \circ }$,
$\therefore \angle BAC = 180 ^ { \circ } - 30 ^ { \circ } - 30 ^ { \circ } = 120 ^ { \circ }$.
$\because \angle DAB = 45 ^ { \circ }$,
$\therefore \angle DAC = \angle BAC - \angle DAB = 120 ^ { \circ } - 45 ^ { \circ } = 75 ^ { \circ }$.
(2)证明:$\because \angle DAB = 45 ^ { \circ }$,$\angle DAC = 75 ^ { \circ }$,
$\therefore \angle ADC = \angle B + \angle DAB = 30 ^ { \circ } + 45 ^ { \circ } = 75 ^ { \circ }$,
$\therefore \angle DAC = \angle ADC$,$\therefore DC = AC$.
$\because AB = AC$,$\therefore DC = AB$.

13.(20分)(2024·海州区期中)在四边形ABDC中,AC= AB,DC= DB,∠CAB= 60°,∠CDB= 120°,E是AC上一点,F是AB延长线上一点,且CE= BF.
(1)在图①中,试说明：DE= DF；
(2)在图②中,若点G在AB上且∠EDG= 60°,试猜想CE,EG,BG之间的数量关系,并证明所归纳结论；
(3)若题中条件“∠CAB= 60°且∠CDB= 120°”改为∠CAB= α,∠CDB= 180°-α,点G在AB上,∠EDG满足什么条件时,(2)中结论仍然成立？

答案： 解:
(1)$\because \angle CAB + \angle C + \angle CDB + \angle ABD = 360 ^ { \circ }$,
$\angle CAB = 60 ^ { \circ }$,$\angle CDB = 120 ^ { \circ }$,
$\therefore \angle C + \angle ABD = 180 ^ { \circ }$.
又$\because \angle DBF + \angle ABD = 180 ^ { \circ }$,$\therefore \angle C = \angle DBF$.
在$\triangle CDE$和$\triangle BDF$中,$\begin{cases}CD = BD\\\angle C = \angle DBF\\CE = BF\end{cases}$
$\therefore \triangle CDE \cong \triangle BDF (SAS)$,$\therefore DE = DF$.
(2)$CE + BG = EG$. 证明:$\because \angle BDC = 120 ^ { \circ }$,$\angle EDG = 60 ^ { \circ }$,
$\therefore \angle EDC + \angle BDG = \angle BDC - \angle EDG = 60 ^ { \circ }$.
由
(1),得$\triangle CDE \cong \triangle BDF$,$\therefore \angle CDE = \angle BDF$,
$\therefore \angle BDG + \angle BDF = 60 ^ { \circ }$,即$\angle FDG = 60 ^ { \circ }$,
$\therefore \angle EDG = \angle FDG$.
在$\triangle DEG$和$\triangle DFG$中,$\begin{cases}DE = DF\\\angle EDG = \angle FDG\\DG = DG\end{cases}$
$\therefore \triangle DEG \cong \triangle DFG (SAS)$,$\therefore EG = FG$.
又$\because CE = BF$,$FG = BF + BG$,$\therefore CE + BG = EG$.
(3)当$\angle EDG = 90 ^ { \circ } - \frac { 1 } { 2 } \alpha$时,$CE + BG = EG$仍然成立,
理由:$\because \angle BDC = 180 ^ { \circ } - \alpha$,$\angle EDG = 90 ^ { \circ } - \frac { 1 } { 2 } \alpha$,
$\therefore \angle EDC + \angle BDG = \angle BDC - \angle EDG = 90 ^ { \circ } - \frac { 1 } { 2 } \alpha$.
由
(1),得$\triangle CDE \cong \triangle BDF$,$\therefore \angle CDE = \angle BDF$,
$\therefore \angle BDG + \angle BDF = 90 ^ { \circ } - \frac { 1 } { 2 } \alpha$,即$\angle FDG = 90 ^ { \circ } - \frac { 1 } { 2 } \alpha$,
$\therefore \angle EDG = \angle FDG$.
在$\triangle DEG$和$\triangle DFG$中,$\begin{cases}DE = DF\\\angle EDG = \angle FDG\\DG = DG\end{cases}$
$\therefore \triangle DEG \cong \triangle DFG (SAS)$,$\therefore EG = FG$.
又$\because CE = BF$,$FG = BF + BG$,$\therefore CE + BG = EG$.
$\therefore$当$\angle EDG = 90 ^ { \circ } - \frac { 1 } { 2 } \alpha$时,$CE + BG = EG$仍然成立.