《2025年点对点期末复习及智胜暑假八年级数学北师大版》

3. 如图,在$\triangle ABC$中,已知$AB = 8$,$BC = 6$,$AC = 7$,依次连接$\triangle ABC$的三边中点,得到$\triangle A_1B_1C_1$,再依次连接$\triangle A_1B_1C_1$的三边中点,得到$\triangle A_2B_2C_2$,…,按这样的规律下去,$\triangle A_{2023}B_{2023}C_{2023}$的周长为$\frac{21}{2^{2023}}$。

答案： 3. $\frac{21}{2^{2023}}$

4. 如图,在$\triangle ABC$中,$CD$是中线,点$E在AB$的延长线上,$AB = AC = BE$,连接$CE$,求证：$CE = 2CD$。

答案：
4. 证明：如图，取$AC$的中点$F$，连接$BF$，

$\because AB = BE$，
$\therefore BF$为$\triangle AEC$的中位线，
$\therefore CE = 2BF$，
$\because AB = AC, D, F$分别为$AB, AC$的中点，
$\therefore AF = AD$，
在$\triangle ABF$与$\triangle ACD$中，$\left\{\begin{array}{l} AB = AC \\ \angle A = \angle A \\ AF = AD \end{array}\right.$，
$\therefore \triangle ABF \cong \triangle ACD(SAS)$，
$\therefore CD = BF, \therefore CE = 2CD$。

5. 如图,在$\triangle ABC$中,$\angle BAC = 90^{\circ}$,$\angle B = 45^{\circ}$,$BC = 10$,过点$A作AD// BC$,且点$D在点A$的右侧。点$P从点A出发沿射线AD$方向以每秒 1 个单位的速度运动,同时点$Q从点C出发沿射线CB$方向以每秒 2 个单位的速度运动,在线段$QC上取点E$,使得$QE = 2$,连接$PE$,设点$P的运动时间为t$秒。
(1) 若$PE\perp BC$,求$BQ$的长；
(2) 是否存在$t$,使以$A$,$B$,$E$,$P$为顶点的四边形为平行四边形？若存在,求出$t$的值；若不存在,请说明理由。

答案：
5. 解：
(1) 如图，作$AM \perp BC$于$M$，记$AC$与$PE$的交点为$N$，

$\because \angle BAC = 90^{\circ}, \angle B = 45^{\circ}, \therefore \angle C = 45^{\circ} = \angle B$，
$\therefore AB = AC, \therefore AM = BM = CM = \frac{1}{2}BC = 5$，
$\because AD // BC, \therefore \angle PAN = \angle C = 45^{\circ}$，
$\because PE \perp BC, \therefore PE = AM = 5, PE \perp AD$，
$\therefore \triangle APN$和$\triangle CEN$都是等腰直角三角形，
$\therefore PN = AP = t, CE = NE = 5 - t$，
$\because CE = CQ - QE = 2t - 2$，
$\therefore 5 - t = 2t - 2$，解得：$t = \frac{7}{3}$，
$\therefore BQ = BC - CQ = 10 - 2 × \frac{7}{3} = \frac{16}{3}$；
(2) 存在，
若以$A, B, E, P$为顶点的四边形为平行四边形，则$AP = BE$，
$\therefore t = 10 - 2t + 2$或$t = 2t - 2 - 10$，解得：$t = 4$或$12$，
$\therefore$ 存在$t$的值，使以$A, B, E, P$为顶点的四边形为平行四边形，$t = 4$或$12$。

6. 如图,线段$AC与BD交于O$,$DO = DC$,$AO = AB$,$E$,$F$,$G分别是OB$,$OC$,$AD$的中点。
(1) 如图 1,当$\angle AOB = 60^{\circ}$时,$EG与FG$的数量关系是______,$\angle EGF = $______；如图 2,当$\angle AOB = 45^{\circ}$时,$EG与FG$的数量关系是______,$\angle EGF = $______；
(2) 如图 3,当$\angle AOB = \theta$时,$EG与FG$的数量关系是______,$\angle EGF = $______；
(3) 请你从上述三个结论中选择一个结论加以证明。(注：直角三角形斜边上的中线是斜边的一半)

答案：
6. 解：
(1) $EG = FG, 60^{\circ}$；$EG = FG, 90^{\circ}$；
(2) $EG = FG, 180^{\circ} - 2\theta$；
(3) ① 当$\angle AOB = 60^{\circ}$时，
证明：如图 1，连接$DF, AE$，

$\because DO = DC, AO = AB, \angle DOC = \angle AOB = 60^{\circ}$，
$\therefore \triangle DOC$与$\triangle AOB$都是等边三角形，
$\because E, F$分别是$OB, OC$的中点，
$\therefore DF \perp AC, AE \perp BD$，
$\because G$是$AD$的中点，$\therefore EG = \frac{1}{2}AD, FG = \frac{1}{2}AD$，
$\therefore EG = FG$，
$\because \angle FDO = \frac{1}{2} \angle CDO = \angle OAE = \frac{1}{2} \angle BAO = 30^{\circ}$，
$\angle ODA + \angle OAD = \angle AOB = 60^{\circ}$，
$\therefore \angle GDF + \angle EAG = 120^{\circ}$，
$\because DG = GF = AG = EG = \frac{1}{2}AD$，
$\therefore \angle DFG = \angle GDF, \angle AEG = \angle EAG$，
$\therefore \angle DFG + \angle AEG = \angle GDF + \angle EAG = 120^{\circ}$，
$\therefore \angle DFG + \angle AEG + \angle GDF + \angle EAG = 240^{\circ}$，
$\therefore \angle DGF + \angle AGE = 360^{\circ} - (\angle DFG + \angle AEG + \angle GDF + \angle EAG) = 120^{\circ}$，
$\therefore \angle EGF = 180^{\circ} - (\angle DGF + \angle AGE) = 60^{\circ}$；
② 当$\angle AOB = 45^{\circ}$时，
证明：如图 2，连接$DF, AE$，
$\because DO = DC, AO = AB, \angle DOC = \angle AOB = 45^{\circ}$，
$\therefore \triangle DOC$与$\triangle AOB$都是等腰直角三角形，
$\because E, F$分别是$OB, OC$的中点，
$\therefore DF \perp AC, AE \perp BD$，
$\because G$是$AD$的中点，$\therefore EG = \frac{1}{2}AD, FG = \frac{1}{2}AD$，
$\therefore EG = FG$，
$\because \angle FDO = \frac{1}{2} \angle CDO = \angle OAE = \frac{1}{2} \angle BAO = 45^{\circ}$，
$\angle ODA + \angle OAD = \angle AOB = 45^{\circ}$，
$\therefore \angle GDF + \angle EAG = 135^{\circ}$，
$\because DG = GF = AG = EG = \frac{1}{2}AD$，
$\therefore \angle DFG = \angle GDF, \angle AEG = \angle EAG$，
$\therefore \angle DFG + \angle AEG = \angle GDF + \angle EAG = 135^{\circ}$，
$\therefore \angle DFG + \angle AEG + \angle GDF + \angle EAG = 270^{\circ}$，
$\therefore \angle DGF + \angle AGE = 360^{\circ} - (\angle DFG + \angle AEG + \angle GDF + \angle EAG) = 90^{\circ}$，
$\therefore \angle EGF = 180^{\circ} - (\angle DGF + \angle AGE) = 90^{\circ}$；

③ 当$\angle AOB = \theta$时，
证明：如图 3，连接$DF, AE$，

$\because DO = DC, AO = AB, \therefore \angle DOC = \angle AOB = \theta$，
$\therefore \triangle DOC$与$\triangle AOB$都是等腰三角形，
$\because E, F$分别是$OB, OC$的中点，
$\therefore DF \perp AC, AE \perp BD$，
$\because G$是$AD$的中点，$\therefore EG = \frac{1}{2}AD, FG = \frac{1}{2}AD$，
$\therefore EG = FG$，
$\because \angle FDO = \angle EAO = 90^{\circ} - \theta, \angle ODA + \angle OAD = \theta$，
$\therefore \angle GDF + \angle EAG = 180^{\circ} - \theta$，
$\because DG = GF = AG = EG = \frac{1}{2}AD$，
$\therefore \angle DFG = \angle GDF, \angle AEG = \angle EAG$，
$\therefore \angle DFG + \angle AEG = \angle GDF + \angle EAG = 180^{\circ} - \theta$，
$\therefore \angle DFG + \angle AEG + \angle GDF + \angle EAG = 360^{\circ} - 2\theta$，
$\therefore \angle DGF + \angle AGE = 360^{\circ} - (\angle DFG + \angle AEG + \angle GDF + \angle EAG) = 2\theta$，
$\therefore \angle EGF = 180^{\circ} - (\angle DGF + \angle AGE) = 180^{\circ} - 2\theta$。