《2025年考出好成绩七年级数学下册鲁教版》

10. 已知：如图，$AD// BC$，点$E$，$F$分别在$DC$，$AB$延长线上，$\angle DCB = \angle DAB$，$AE\perp EF$.
(1) 求证：$DC// AB$；
(2) 若$\angle DEA = 30^{\circ}$，求$\angle AFE$的大小.

答案： 解：
(1) 证明：$\because AD// BC$
$\therefore\angle ABC+\angle DAB = 180^{\circ}$
$\because\angle DCB=\angle DAB$
$\therefore\angle ABC+\angle DCB = 180^{\circ}$
$\therefore DC// AB$
(2) 由
(1)知$DC// AB$
$\therefore\angle EAF=\angle DEA = 30^{\circ}$
$\because AE\perp EF,\therefore\angle AEF = 90^{\circ}$
$\therefore\angle AFE = 90^{\circ}-\angle EAF = 60^{\circ}$

11. 如图，在$\triangle ABC$中，点$D$，$E$分别在边$AB$，$AC$上，将$\triangle ADE$沿$DE$折叠至$\triangle FDE$位置，点$A$的对应点为$F$. 若$\angle A = 15^{\circ}$，$\angle BDF = 120^{\circ}$，则$\angle CEF$的度数为（ ）

A. $90^{\circ}$
B. $100^{\circ}$
C. $110^{\circ}$
D. $120^{\circ}$

答案： A

12. 如图，点$D$，$E$分别在$\triangle ABC$的边$AB$，$AC$上，点$F$在线段$CD$上，且$\angle 1 = \angle B$，$DE// BC$.
(1) 求证：$AB// EF$；
(2) 若$DE$平分$\angle ADC$，$\angle ACB = 2\angle B$，求证：$\angle DFE = \angle DEC$.

答案： 证明：
(1)$\because DE// BC$
$\therefore\angle ADE=\angle B$
$\because\angle 1=\angle B,\therefore\angle ADE=\angle 1$
$\therefore AB// EF$
(2) 由
(1)知，$\angle ADE=\angle 1=\angle B$
$\because DE$平分$\angle ADC$
$\therefore\angle ADE=\angle CDE$
$\therefore\angle ADE=\angle CDE=\angle 1=\angle B$
$\because DE// BC,\therefore\angle CDE=\angle BCD$
$\therefore\angle BCD=\angle 1=\angle B$
$\because\angle ACB = 2\angle B,\angle ACB=\angle BCD+\angle ACD$
$\therefore\angle BCD+\angle ACD = 2\angle B$
$\therefore\angle ACD=\angle B=\angle 1$
$\therefore\angle DFE=\angle ACD+\angle CEF=\angle 1+\angle CEF=\angle DEC$

13. 推理能力 已知：直线$EF$分别交直线$AB$，$CD$于点$G$，$H$，$\angle AGE+\angle CHF = 180^{\circ}$.
(1) 如图1，求证：$AB// CD$；
(2) 如图2，$N$为直线$AB$，$CD$之间的一点，$\angle AGH = 2\angle CHG$，$\angle AGN:\angle NGH = 3:2$，$\angle GHN:\angle CHN = 1:2$，求$\angle N$的度数；
(3) 如图3，$M$，$N$分别为直线$AB$，$CD$之间不同的两点，连接$GM$，$HM$，$GN$，$HN$，且$GN$平分$\angle AGM$，$HN$平分$\angle CHM$，$\angle M-\angle N = 30^{\circ}$，求$\angle N$的度数.

答案： 解：
(1) 证明：$\because\angle AGE+\angle AGH = 180^{\circ},\angle AGE+\angle CHF = 180^{\circ}$
$\therefore\angle AGH=\angle CHF,\therefore AB// CD$
(2) 由
(1)可知，$AB// CD$
$\therefore\angle AGH+\angle CHG = 180^{\circ}$
$\because\angle AGH = 2\angle CHG$
$\therefore\angle AGH = 120^{\circ},\angle CHG = 60^{\circ}$
$\because\angle AGN:\angle NGH = 3:2,\angle AGN+\angle NGH = 120^{\circ}$
$\therefore\angle AGN = 72^{\circ},\angle NGH = 48^{\circ}$
$\because\angle GHN:\angle CHN = 1:2,\angle GHN+\angle CHN = 60^{\circ}$
$\therefore\angle GHN = 20^{\circ},\angle CHN = 40^{\circ}$
$\therefore\angle N = 180^{\circ}-\angle NGH-\angle GHN = 180^{\circ}-48^{\circ}-20^{\circ}=112^{\circ}$
(3) 由
(2)可知，$\angle N=\angle AGN+\angle CHN$
如图，过点$M$作$MQ// AB$，交$GH$于点$Q$
$\because AB// CD$
$\therefore MQ// AB// CD$
$\therefore\angle AGM+\angle GMQ = 180^{\circ},\angle QMH+\angle CHM = 180^{\circ}$
$\therefore\angle GMH = 360^{\circ}-\angle AGM-\angle CHM$
$\because GN$平分$\angle AGM$，$HN$平分$\angle CHM$
$\therefore\angle AGN=\frac{1}{2}\angle AGM,\angle CHN=\frac{1}{2}\angle CHM$
$\therefore\angle N=\frac{1}{2}(\angle AGM+\angle CHM)$
$\therefore\angle GMH = 360^{\circ}-2\angle N$
$\because\angle GMH-\angle N = 30^{\circ}$
$\therefore\angle N = 110^{\circ}$