Solution: (1)① In Rt△ABC, D is the midpoint of AB,

∴AD=BD=AD=，B=∠BDC=60

∫∠A = 30，

∴∠ACD=60 -30 =30，

When ∵∠CDE = 60° or ∠ CDF = 60,

∴∠CKD=90，

△ ∴ in △CDA, AM(K)=CM(K), that is, AM(K)=KM(C) (the vertical line of the base of an isosceles triangle coincides with the center line).

∵CK=0 or AM=0,

∴am+ck=mk； (2 points)

(2) The same is true for (1)

∠ACD=30，∠CDB=60，

∠∠A = 30，∠CDF=30，∠ EDF = 60，

∴∠ADM=30，

∴AM=MD,CK=KD，

∴AM+CK=MD+KD，

△MKD, am+CK > MK in MK (the sum of two sides is greater than the third side). (2 points)

(2) > (2 points)

Prove that the symmetric point g of point c is about FD,

Connect GK, GM, Global,

Then CD=GD, GK=CK, ∠GDK=∠CDK,

D is the midpoint of ∴ad=cd=gd ab,

∫∠a = 30 ,∴∠cda= 120，

∫∠EDF = 60 ,∴∠gdm+∠gdk=60，

∠ADM+∠CDK=60。

∴∠adm=∠gdm(3 points)

DM = DM，

∴△ADM≌△GDM,∴GM=AM.

∫GM+GK > MK，∴ AM+CK > MK. ( 1)

(3) solution: from (2), GM=AM, GK=CK,

∫MK2+CK2 = AM2，

∴MK2+GK2=GM2，

∴∠GKM=90，

There is also the symmetrical point g of point ∵c about FD,

∴