證明: (1)$\because AB = AC，$$\angle BAC = 120^{\circ}，$$D$是$BC$的中點(diǎn)，$\therefore \angle BAD = \angle CAD=\frac{1}{2}\angle BAC = 60^{\circ}，$即$AD$平分$\angle BAC。$$\because DM\perp AB，$$\therefore \angle AMD = 90^{\circ}，$$\therefore \angle ADM = 90^{\circ}-\angle MAD = 90^{\circ}-60^{\circ}=30^{\circ}。$$\because \angle MDN = 60^{\circ}，$$\therefore \angle ADN = \angle MDN - \angle MDA = 60^{\circ}-30^{\circ}=30^{\circ}，$$\therefore \angle AND = 180^{\circ}-\angle NAD - \angle ADN = 90^{\circ}，$$\therefore DN\perp AC。$$\because AD$平分$\angle BAC，$$\therefore DM = DN。$

(2)相等，理由如下：如圖①，取$AB$的中點(diǎn)$E，$連接$DE，$$\because AB = AC，$$\angle BAC = 120^{\circ}，$$D$是$BC$的中點(diǎn)，$\therefore AD\perp BC，$$\angle BAD = \angle CAD = 60^{\circ}，$$\therefore \angle ADB = 90^{\circ}，$$\therefore DE = AE = BE，$$\therefore \triangle AED$是等邊三角形，$\therefore DE = AD，$$\angle AED = \angle ADE = 60^{\circ}。$$\because \angle MDN = 60^{\circ}，$$\therefore \angle EDM + \angle MDA = \angle ADN + \angle MDA，$$\therefore \angle EDM = \angle ADN。$在$\triangle EDM$和$\triangle ADN$中，$\begin{cases}\angle MED = \angle NAD \\ DE = AD \\ \angle EDM = \angle ADN\end{cases}，$$\therefore \triangle EDM\cong\triangle ADN(ASA)，$$\therefore DM = DN。$

(3)$BM + NC=\frac{3}{2}AC。$

證明：分兩種情況討論：

①如圖①，當(dāng)點(diǎn)$N$在線段$AC$上時(shí)，由

(2)可知$\triangle EDM\cong\triangle ADN，$$\therefore EM = AN。$$\because AM + EM = AE=\frac{1}{2}AB，$$\therefore AM + AN=\frac{1}{2}AB。$$\because AB = AC，$$\therefore (AC - BM)+(AC - NC)=\frac{1}{2}AC，$$\therefore BM + NC=\frac{3}{2}AC。$

②如圖②，當(dāng)點(diǎn)$N$在$CA$的延長線上時(shí)，由

(2)可知$\triangle ADE$是等邊三角形，$\angle AED = \angle ADE = 60^{\circ}。$$\because \angle MDN = 60^{\circ}，$$\therefore \angle MDN - \angle NDE = \angle ADE - \angle NDE，$即$\angle EDM = \angle ADN。$$\because \angle EMD + \angle EDM = \angle AED = 60^{\circ}，$$\angle N + \angle ADN = \angle CAD = 60^{\circ}，$$\therefore \angle EMD = \angle N。$在$\triangle ADN$和$\triangle EDM$中，$\begin{cases}\angle N = \angle EMD \\ \angle ADN = \angle EDM \\ AD = ED\end{cases}，$$\therefore \triangle ADN\cong\triangle EDM(AAS)，$$\therefore AN = EM。$$\because BM + ME = BE=\frac{1}{2}AB=\frac{1}{2}AC，$$\therefore BM + AN=\frac{1}{2}AC，$$\therefore BM + AN + AC=\frac{3}{2}AC，$即$BM + NC=\frac{3}{2}AC。$

綜上所述，$BM + NC=\frac{3}{2}AC。$

證明: (1)如圖①，作$CM\perp BD$于點(diǎn)$M，$$CN\perp AD，$交$AD$的延長線于點(diǎn)$N。$$\because \triangle ABC$是等邊三角形，$\therefore AC = BC，$$\angle BCA = \angle ADB = 60^{\circ}。$$\because \angle BGC = \angle AGD，$$\therefore \angle CBM = \angle CAD。$在$\triangle BCM$和$\triangle ACN$中，$\begin{cases}\angle BMC = \angle ANC \\ \angle CBM = \angle CAN \\ BC = AC\end{cases}，$$\therefore \triangle BCM\cong\triangle ACN(AAS)，$$\therefore CM = CN。$又$\because CM\perp BD，$$CN\perp AD，$$\therefore \angle MDC = \angle NDC = 60^{\circ}，$$\therefore \angle ADB = \angle ADF = 60^{\circ}，$$\therefore DA$平分$\angle BDF。$

(2)設(shè)$\angle ABD = \alpha，$則$\angle ACD = 2\alpha，$$\therefore \angle BCD = 60^{\circ}+2\alpha，$$\therefore \angle BDC = 60^{\circ}-\alpha=\angle DBC，$$\therefore BC = DC，$$\angle F = 60^{\circ}-\alpha-\alpha = 60^{\circ}-2\alpha。$$\because FH = FC，$$\therefore \angle FHC = \angle FCH = 60^{\circ}+\alpha。$$\therefore \angle HCB = \angle BCD - \angle FCH=\alpha。$$\therefore \angle HCB = \angle ABD。$在$\triangle ABG$和$\triangle BCH$中，$\begin{cases}\angle BAG = \angle CBH \\ AB = BC \\ \angle ABG = \angle BCH\end{cases}，$$\therefore \triangle ABG\cong\triangle BCH(ASA)，$$\therefore BH = AG。$

(3)由

(2)知$\angle ACK = 60^{\circ}-2\alpha，$$\angle KGC = \angle BDC + \angle ACD = 60^{\circ}+\alpha，$$\therefore$在$\triangle GCK$中，$\angle GKC = 180^{\circ}-\angle ACK - \angle KGC = 60^{\circ}+\alpha=\angle KGC。$$\therefore CG = CK。$$\because \triangle BCH$翻折得到$\triangle NCH，$$\therefore CN = BC = AC = AB，$$HN = HB，$$\angle N = \angle ABC = 60^{\circ}。$$\because CN - CK = CA - CG，$即$NK = AG。$如圖②，連接$HK，$$\therefore BH = HN = NK = AG，$$\therefore \triangle NHK$是等邊三角形，$\therefore HK = KN = BH，$$\therefore \angle HBK = \angle HKB。$在$BK$上截取$BP = KM，$在$\triangle BHP$和$\triangle KHM$中，$\begin{cases}BH = KH \\ \angle HBP = \angle HKM \\ BP = KM\end{cases}，$$\therefore \triangle BHP\cong\triangle KHM(SAS)，$$\therefore HP = HM，$$BP = KM = 6，$$\therefore PM = 8 - 6 = 2。$$\because \angle DMH = \angle BDC + \angle DCH，$$\therefore \angle DMH = 60^{\circ}-\alpha+2\alpha+60^{\circ}-\alpha = 120^{\circ}，$$\therefore \angle HMB = 60^{\circ}，$$\therefore \triangle HMP$是等邊三角形，$\therefore HM = PM = 2。$