The quadrilateral ABCD is a diamond,
∴AC⊥BD, BD and so on ∠ ADC. Ao = DC = BC,
∴∠COD=∠COB=∠AOD=90。
∠ADO=∠ADC=×60 =30，
And f are the midpoint of DC and CB, respectively.
∴OE=CD,OF=BC,AO=AD，
∴0E=OF=OA，
Point o is the outer center of △AEF.
(2) Solution:
① conjecture: the epicenter p must fall on the straight line DB.
Proof: As shown in Figure 2, connect PE and PA respectively, and make PI⊥CD in I and PJ⊥AD in J through point P respectively.
∴∠PIE=∠PJD=90，
∫∠ADC = 60，
∴∠ipj=360-∠pie-∠pjd-∠jdi = 120，
Point p is the outer center of equilateral △AEF,
∴∠EPA= 120，PE=PA，
∴∠IPJ=∠EPA，
∴∠IPE=∠JPA，
∴△PIE≌△PJA，
∴PI=PJ，
Point p is on the bisector of ∠ADC, that is, point p falls on the straight line DB.
② 1/DM+ 1/DN is a fixed value of 2.
When AE⊥DC, the area of △ AEF is the smallest.
At this time, point E and point F are the midpoint of DC and CB respectively.
Connect BD and AC at point P, which is defined by (1).
The available point p is the outer center of △AEF.
As shown in fig. 3, let MN pass through BC at point G,
Let DM=x and DN = y (x ≠ 0.y ≠ o), then CN=y- 1,
∫BC∨DA，
∴△GBP≌△MDP.
∴BG=DM=x.
∴CG= 1-x
? ∫BC∨DA，
∴△NCG∽△NDM，
∴CN/DN=CG/DM，
∴(y- 1)/y=( 1-x)/x，
∴x+y=2xy，
∴ 1/x+ 1/y=2，
That is 1/DM+ 1/DN = 2.