22. (10分)(2025·漳州期末)已知在$△ABC$中,$AB= AC,∠BAC= 120^{\circ}$,D是BC的中点,M是AB边上的一点,连接DM,作$∠MDN= 60^{\circ}$交直线AC于点N.
(1)如图①,当$DM⊥AB$时,求证:$DM= DN$.
(2)如图②,当M是AB边上任意一点时,DM与DN还相等吗? 请说明理由.
(3)请写出AC,BM,NC之间的数量关系,并证明.

(1)∵AB = AC,∠BAC = 120°,D是BC的中点,∴∠BAD = ∠CAD = $\frac{1}{2}$∠BAC = 60°,即AD平分∠BAC.∵DM⊥AB,∴∠AMD = 90°,∴∠ADM = 90° - ∠MAD = 90° - 60° = 30°.∵∠MDN = 60°,∴∠ADN = ∠MDN - ∠MDA = 60° - 30° = 30°,∴∠AND = 180° - ∠NAD - ∠ADN = 90°,∴DN⊥AC.∴AD平分∠BAC,∴DM = DN.
(2)相等,理由如下:如图①,取AB的中点E,连接DE,∵AB = AC,∠BAC = 120°,D是BC的中点,∴AD⊥BC,∠BAD = ∠CAD = 60°,∴∠ADB = 90°,∴DE = AE = BE,∴△AED是等边三角形,∴DE = AD,∠AED = ∠ADE = 60°.∵∠MDN = 60°,∴∠EDM + ∠MDA = ∠ADN + ∠MDA,∴∠EDM = ∠ADN.在△EDM和△ADN中,{∠MED = ∠NAD,DE = AD,∠EDM = ∠ADN,∴△EDM≌△ADN(ASA),∴DM = DN.

(3)BM + NC = $\frac{3}{2}$AC.
证明:分两种情况讨论:
①如图①,当点N在线段AC上时,由(2)可知△EDM≌△ADN,∴EM = AN.∵AM + EM = AE = $\frac{1}{2}$AB,∴AM + AN = $\frac{1}{2}$AB.∵AB = AC,∴(AC - BM) + (AC - NC) = $\frac{1}{2}$AC,∴BM + NC = $\frac{3}{2}$AC.
②如图②,当点N在CA的延长线上时,由(2)可知△ADE是等边三角形,∠AED = ∠ADE = 60°.∵∠MDN = 60°,∴∠MDN - ∠NDE = ∠ADE - ∠NDE,即∠EDM = ∠ADN.∵∠EMD + ∠EDM = ∠AED = 60°,∠N + ∠ADN = ∠CAD = 60°,∴∠EMD = ∠N.在△ADN和△EDM中,{∠N = ∠EMD,∠ADN = ∠EDM,AD = ED,∴△ADN≌△EDM(AAS),∴AN = EM.∵BM + ME = BE = $\frac{1}{2}$AB = $\frac{1}{2}$AC,∴BM + AN = $\frac{1}{2}$AC,∴BM + AN + AC = $\frac{3}{2}$AC,即BM + NC = $\frac{3}{2}$AC.
综上所述,BM + NC = $\frac{3}{2}$AC.

23. (12分)已知$△ABC$是等边三角形,点D是

射线CF上一点,连接BD交线段AC于点G.
(1)如图①,当$∠ADB= 60^{\circ}$时,求证:DA平分$∠BDF$;
(2)如图②,延长BA交射线CF于点F,当$∠ACD= 2∠ABD$时,在AB上取一点H,且$FH= FC$,连接CH,求证:$BH= AG$;
(3)如图③,在(2)的条件下,将$△BCH$沿CH翻折,得到$△NHC$,CH与BD交于点M,CN与BD交于点K,若$BM= 8,MK= 6$,求HM的长.

(1)如图①,作CM⊥BD于点M,CN⊥AD,交AD的延长线于点N.∵△ABC是等边三角形,∴AC = BC,∠BCA = ∠ADB = 60°.∵∠BGC = ∠AGD,∴∠CBM = ∠CAD.在△BCM和△ACN中,{∠BMC = ∠ANC,∠CBM = ∠CAN,BC = AC,∴△BCM≌△ACN(AAS),∴CM = CN.又∵CM⊥BD,CN⊥AD,∴∠MDC = ∠NDC = 60°,∴∠ADB = ∠ADF = 60°,∴DA平分∠BDF.

(2)设∠ABD = α,则∠ACD = 2α,∴∠BCD = 60° + 2α,∴∠BDC = 60° - α = ∠DBC,∴BC = DC,∠F = 60° - α - α = 60° - 2α.∵FH = FC,∴∠FHC = ∠FCH = 60° + α.∴∠HCB = ∠BCD - ∠FCH = α.∴∠HCB = ∠ABD.在△ABG和△BCH中,{∠BAG = ∠CBH,AB = BC,∠ABG = ∠BCH,∴△ABG≌△BCH(ASA),∴BH = AG.
(3)由(2)知∠ACK = 60° - 2α,∠KGC = ∠BDC + ∠ACD = 60° + α,∴在△GCK中,∠GKC = 180° - ∠ACK - ∠KGC = 60° + α = ∠KGC.∴CG = CK.∵△BCH翻折得到△NCH,∴CN = BC = AC = AB,HN = HB,∠N = ∠ABC = 60°.∵CN - CK = CA - CG,即NK = AG.如图②,连接HK,∵BH = HN = NK = AG,∴△NHK是等边三角形,∴HK = KN = BH,∴∠HBK = ∠HKB.在BK上截取BP = KM,在△BHP和△KHM中,{BH = KH,∠HBP = ∠HKM,BP = KM,∴△BHP≌△KHM(SAS),∴HP = HM,BP = KM = 6,∴PM = 8 - 6 = 2.∵∠DMH = ∠BDC + ∠DCH,∴∠DMH = 60° - α + 2α + 60° - α = 120°,∴∠HMB = 60°,∴△HMP是等边三角形,∴HM = PM = 2.