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7.(2024·高新区期中)如图，在$\triangle ABC$中，$AB = AC$，$\angle B = 30^{\circ}$，$AD \perp AB$，交$BC$于点$D$。若$AD = 1$，则$CD$的长为
1
。答案
7. 1
解析
解：
∵ $AB = AC$，$\angle B = 30°$，
∴ $\angle C = \angle B = 30°$，$\angle BAC = 180° - 30° - 30° = 120°$。
∵ $AD \perp AB$，
∴ $\angle BAD = 90°$，$\angle CAD = \angle BAC - \angle BAD = 120° - 90° = 30°$。
在 $Rt\triangle ABD$ 中，$\angle B = 30°$，$AD = 1$，
∴ $BD = 2AD = 2$，$\angle ADB = 60°$。
∵ $\angle ADB = \angle C + \angle CAD$，$\angle C = 30°$，$\angle CAD = 30°$，
∴ $\angle ADB = 60°$，故 $\triangle ADC$ 中，$\angle CAD = \angle C = 30°$，
∴ $CD = AD = 1$。
1
∵ $AB = AC$，$\angle B = 30°$，
∴ $\angle C = \angle B = 30°$，$\angle BAC = 180° - 30° - 30° = 120°$。
∵ $AD \perp AB$，
∴ $\angle BAD = 90°$，$\angle CAD = \angle BAC - \angle BAD = 120° - 90° = 30°$。
在 $Rt\triangle ABD$ 中，$\angle B = 30°$，$AD = 1$，
∴ $BD = 2AD = 2$，$\angle ADB = 60°$。
∵ $\angle ADB = \angle C + \angle CAD$，$\angle C = 30°$，$\angle CAD = 30°$，
∴ $\angle ADB = 60°$，故 $\triangle ADC$ 中，$\angle CAD = \angle C = 30°$，
∴ $CD = AD = 1$。
1
8.（2024·吴江区期中）如图，在$\triangle ABC$中，$AB = AC$，$D$是$BC$边的中点，连接$AD$，$BM$平分$\angle ABC$，交$AC$于点$M$，过点$M$作$MN // BC$，交$AB$于点$N$。
（1）若$\angle C = 72^{\circ}$，求$\angle BAD$的度数；
（2）求证：$NB = NM$。
（1）若$\angle C = 72^{\circ}$，求$\angle BAD$的度数；
（2）求证：$NB = NM$。
答案
8. （1）$\because AB = AC$，$D$是$BC$边的中点，$\therefore AD\perp BC$，$\angle BAD = \angle CAD$，$\therefore \angle ADC = 90^{\circ}.\because \triangle ADC$的内角和为$180^{\circ},\angle C = 72^{\circ},\therefore \angle CAD = 180^{\circ} - \angle ADC - \angle C = 18^{\circ}$，$\therefore \angle BAD = 18^{\circ}$ （2）$\because MN// BC$，$\therefore \angle NMB = \angle CBM$.$\because BM$平分$\angle ABC$，$\therefore \angle NBM = \angle CBM$，$\therefore \angle NBM = \angle NMB$，$\therefore NB = NM$
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，$\therefore AN\perp CM$，$\therefore \angle ANM = 90^{\circ}$，$\therefore \angle ANM + \angle CME = 180^{\circ}$，$\therefore AN// EM$
