无码精品人妻一区二区三区湄公河 ,无码精品人妻一区二区三区湄公河

A

$8$

$6$或$12$

$4$

證明: (1)$\because \angle A = 30^{\circ}，$$AD = AC，$$\therefore \angle ADC=\angle ACD=\dfrac{1}{2}\times(180^{\circ}-30^{\circ}) = 75^{\circ}.$又$\because \angle ACB = 90^{\circ}，$$\therefore \angle ABC = 60^{\circ}.$$\because BE$平分$\angle ABC，$$\therefore \angle ABE = 30^{\circ}，$$\therefore \angle DMB=\angle ADC-\angle ABE = 75^{\circ}-30^{\circ}=45^{\circ}；$

 (2)$\because \angle ABE=\angle CBE = 30^{\circ}，$$\therefore BE = 2CE = 2.$$\because \angle ABE=\angle A = 30^{\circ}，$$\therefore AE = BE = 2，$$\therefore AC = 3，$$\therefore AD = AC = 3；$

 (3)$\because CH\perp BE，$$\therefore \angle CHB = 90^{\circ}.$$\because \angle HMC=\angle DMB = 45^{\circ}，$$\therefore \angle HCM=\angle HMC = 45^{\circ}，$$\therefore HM = HC.$$\because \angle CHB = 90^{\circ}，$$\angle EBC=\dfrac{1}{2}\angle ABC = 30^{\circ}，$$\therefore HC = HM=\dfrac{1}{2}BC.$$\because \angle A = 30^{\circ}，$$\angle ACB = 90^{\circ}，$$\therefore BC=\dfrac{1}{2}AB，$$\therefore AB = 2BC = 4MH.$

證明: (1)如圖，過點$M$作$MQ// BC，$交$AC$于點$Q，$在等邊$\triangle ABC$中，$\angle A=\angle B=\angle ACB = 60^{\circ}.$$\because MQ// BC，$$\therefore \angle AMQ=\angle B = 60^{\circ}，$$\angle AQM=\angle ACB = 60^{\circ}，$$\angle QMP=\angle N，$$\therefore \triangle AMQ$是等邊三角形，$\therefore AM = QM.$$\because AM = CN，$$\therefore QM = CN.$在$\triangle QMP$和$\triangle CNP$中，$\begin{cases}\angle QPM=\angle CPN \\ \angle QMP=\angle N \\ QM = CN\end{cases}，$$\therefore \triangle QMP\cong \triangle CNP(AAS)，$$\therefore MP = NP；$

 (2)$\because \triangle AMQ$是等邊三角形，且$MH\perp AC，$$\therefore AH = HQ.$$\because \triangle QMP\cong \triangle CNP，$$\therefore QP = CP，$$\therefore PH = HQ + QP=\dfrac{1}{2}AC.$$\because AB = a，$$AB = AC，$$\therefore PH=\dfrac{1}{2}a.$

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