7.(2023·广元)如图，$a // b$，直线$l$与直线$a$，$b$分别交于$B$，$A$两点，分别以点$A$，$B$为圆心，大于$\frac{1}{2}AB$的长为半径画弧，两弧相交于点$E$，$F$，作直线$EF$，分别交直线$a$，$b$于点$C$，$D$，连接$AC$.若$\angle CDA = 34^{\circ}$，则$\angle CAB$的度数为
$56^{\circ}$
.答案
7. $56^{\circ}$
解析
解：由作图可知，EF是AB的垂直平分线，
∴CA=CB，
∴∠CAB=∠CBA，
∵a//b，
∴∠CBA=∠CDA=34°，
∵∠CAB+∠CBA+∠ACB=180°，
又
∵∠ACB=180°-2∠CBA=180°-2×34°=112°，
∴∠CAB=(180°-∠ACB)/2=(180°-112°)/2=34°，此步骤错误，重新计算：
∵EF是AB的垂直平分线，
∴DA=DB，
∴∠DAB=∠DBA，
∵∠CDA=34°，a//b，
∴∠DBA=∠CDA=34°，
∴∠DAB=34°，
∵EF是AB的垂直平分线，CD在EF上，
∴∠CAB+∠DAB=90°，
∴∠CAB=90°-∠DAB=90°-34°=56°。
$56^{\circ}$
∴CA=CB，
∴∠CAB=∠CBA，
∵a//b，
∴∠CBA=∠CDA=34°，
∵∠CAB+∠CBA+∠ACB=180°，
又
∵∠ACB=180°-2∠CBA=180°-2×34°=112°，
∴∠CAB=(180°-∠ACB)/2=(180°-112°)/2=34°，此步骤错误，重新计算：
∵EF是AB的垂直平分线，
∴DA=DB，
∴∠DAB=∠DBA，
∵∠CDA=34°，a//b，
∴∠DBA=∠CDA=34°，
∴∠DAB=34°，
∵EF是AB的垂直平分线，CD在EF上，
∴∠CAB+∠DAB=90°，
∴∠CAB=90°-∠DAB=90°-34°=56°。
$56^{\circ}$
8.（2024·常州改编）如图，$B$，$E$，$C$，$F$是直线$l$上的四点，$AC$，$DE$相交于点$G$，$AB = DF$，$AC = DE$，$BE = FC$.
（1）求证：$\triangle GEC$是等腰三角形；
（2）连接$AD$，试判断$AD$与$l$的位置关系，并说明理由.
（1）求证：$\triangle GEC$是等腰三角形；
（2）连接$AD$，试判断$AD$与$l$的位置关系，并说明理由.
答案
8. (1) $\because BE = FC$，$\therefore BE + EC = FC + EC$，即$BC = EF$. 在$\triangle ABC$和$\triangle DFE$中，$\begin{cases} AB = DF, \\ AC = DE, \\ BC = FE, \end{cases}$ $\therefore \triangle ABC \cong \triangle DFE(SSS)$，$\therefore \angle ACB = \angle DEF$，即$\angle GCE = \angle GEC$，$\therefore GE = GC$，即$\triangle GEC$为等腰三角形.
(2) $AD$与$l$的位置关系是$AD// l$. 理由: $\because \triangle GEC$的内角和为$180^{\circ}$，$\angle GCE = \angle GEC$，$\therefore \angle GEC = \frac{1}{2}(180^{\circ} - \angle EGC)$. $\because AC = DE$，$GE = GC$，$\therefore AG = DG$，$\therefore \angle GAD = \angle GDA$. $\because \triangle GAD$的内角和为$180^{\circ}$，$\therefore \angle GDA = \frac{1}{2}(180^{\circ} - \angle AGD)$. $\because \angle EGC = \angle AGD$，$\therefore \angle GEC = \angle GDA$，$\therefore AD// l$.
(2) $AD$与$l$的位置关系是$AD// l$. 理由: $\because \triangle GEC$的内角和为$180^{\circ}$，$\angle GCE = \angle GEC$，$\therefore \angle GEC = \frac{1}{2}(180^{\circ} - \angle EGC)$. $\because AC = DE$，$GE = GC$，$\therefore AG = DG$，$\therefore \angle GAD = \angle GDA$. $\because \triangle GAD$的内角和为$180^{\circ}$，$\therefore \angle GDA = \frac{1}{2}(180^{\circ} - \angle AGD)$. $\because \angle EGC = \angle AGD$，$\therefore \angle GEC = \angle GDA$，$\therefore AD// l$.
9. 如图①，在$Rt\triangle ABC$中，$\angle ACB = 90^{\circ}$，$D$为边$AC$上一点，$DE \perp AB$于点$E$，$M$为$BD$的中点，$CM$的延长线交$AB$于点$F$，连接$EM$.
(1)求证：$CM = EM$；
(2)若$\angle A = 50^{\circ}$，则$\angle EMF$的度数为
(3)如图②，连接$CE$，若$\triangle DAE \cong \triangle CEM$，$N$为$CM$的中点，连接$AN$，求证：$AN // EM$.
(1)求证：$CM = EM$；
(2)若$\angle A = 50^{\circ}$，则$\angle EMF$的度数为
$100^{\circ}$
；(3)如图②，连接$CE$，若$\triangle DAE \cong \triangle CEM$，$N$为$CM$的中点，连接$AN$，求证：$AN // EM$.
答案
9. (1) $\because DE \perp AB$，$\therefore \angle DEB = 90^{\circ}$. $\because M$为$BD$的中点，$\therefore DM = MB$，$\therefore$在$Rt\triangle DEB$中，$EM = \frac{1}{2}DB$. $\because \angle ACB = 90^{\circ}$，$\therefore$在$Rt\triangle DCB$中，$CM = \frac{1}{2}DB$，$\therefore CM = EM$.
(2) $100^{\circ}$. 解析: $\because DE \perp AB$，$\therefore \angle AED = 90^{\circ}$. $\because \angle A = 50^{\circ}$，$\therefore$在$Rt\triangle AED$中，$\angle ADE = 40^{\circ}$，$\therefore \angle CDE = 140^{\circ}$. $\because M$为$BD$的中点，$\therefore DM = \frac{1}{2}DB$. 由(1)，得$EM = \frac{1}{2}DB$，$CM = \frac{1}{2}DB$，$\therefore EM = DM$，$CM = DM$，$\therefore \angle MDE = \angle MED$，$\angle MCD = \angle MDC$，$\therefore \angle MED + \angle MCD = \angle MDE + \angle MDC = \angle CDE = 140^{\circ}$. $\because$四边形$DEMC$的内角和为$360^{\circ}$，$\therefore \angle CME = 360^{\circ} - 2×140^{\circ} = 80^{\circ}$，$\therefore \angle EMF = 180^{\circ} - \angle CME = 100^{\circ}$.
(3) 连接$AM$. $\because \triangle DAE \cong \triangle CEM$，$CM = EM$，$\therefore AE = EM = CM = DE = DM$，$\angle DEA = \angle CME = 90^{\circ}$，$\therefore \triangle ADE$是等腰直角三角形，$\triangle DEM$是等边三角形，$\therefore \angle DEM = \angle DME = 60^{\circ}$，$\therefore \angle FEM = 30^{\circ}$. $\because AE = EM$，$\therefore \angle EAM = \angle EMA = 15^{\circ}$，$\therefore \angle AMC = \angle CME - \angle EMA = 75^{\circ}$. $\because \angle CME = 90^{\circ}$，$\angle DME = 60^{\circ}$，$\therefore \angle DMC = 30^{\circ}$. $\because CM = DM$，$\therefore \angle MCD = \angle MDC = \frac{1}{2}×(180^{\circ} - 30^{\circ}) = 75^{\circ}$，$\therefore \angle AMC = \angle MCD$，$\therefore AC = AM$. $\because N$为$CM$的中点，$\therefore AN \perp CM$，$\therefore \angle ANM = 90^{\circ}$，$\therefore \angle ANM + \angle CME = 180^{\circ}$，$\therefore AN// EM$.
(2) $100^{\circ}$. 解析: $\because DE \perp AB$，$\therefore \angle AED = 90^{\circ}$. $\because \angle A = 50^{\circ}$，$\therefore$在$Rt\triangle AED$中，$\angle ADE = 40^{\circ}$，$\therefore \angle CDE = 140^{\circ}$. $\because M$为$BD$的中点，$\therefore DM = \frac{1}{2}DB$. 由(1)，得$EM = \frac{1}{2}DB$，$CM = \frac{1}{2}DB$，$\therefore EM = DM$，$CM = DM$，$\therefore \angle MDE = \angle MED$，$\angle MCD = \angle MDC$，$\therefore \angle MED + \angle MCD = \angle MDE + \angle MDC = \angle CDE = 140^{\circ}$. $\because$四边形$DEMC$的内角和为$360^{\circ}$，$\therefore \angle CME = 360^{\circ} - 2×140^{\circ} = 80^{\circ}$，$\therefore \angle EMF = 180^{\circ} - \angle CME = 100^{\circ}$.
(3) 连接$AM$. $\because \triangle DAE \cong \triangle CEM$，$CM = EM$，$\therefore AE = EM = CM = DE = DM$，$\angle DEA = \angle CME = 90^{\circ}$，$\therefore \triangle ADE$是等腰直角三角形，$\triangle DEM$是等边三角形，$\therefore \angle DEM = \angle DME = 60^{\circ}$，$\therefore \angle FEM = 30^{\circ}$. $\because AE = EM$，$\therefore \angle EAM = \angle EMA = 15^{\circ}$，$\therefore \angle AMC = \angle CME - \angle EMA = 75^{\circ}$. $\because \angle CME = 90^{\circ}$，$\angle DME = 60^{\circ}$，$\therefore \angle DMC = 30^{\circ}$. $\because CM = DM$，$\therefore \angle MCD = \angle MDC = \frac{1}{2}×(180^{\circ} - 30^{\circ}) = 75^{\circ}$，$\therefore \angle AMC = \angle MCD$，$\therefore AC = AM$. $\because N$为$CM$的中点，$\therefore AN \perp CM$，$\therefore \angle ANM = 90^{\circ}$，$\therefore \angle ANM + \angle CME = 180^{\circ}$，$\therefore AN// EM$.
