Solution: (1) Solution: Therefore, the answer is: the isosceles triangle has three lines in one (or the bisector of the top corner of the isosceles triangle, the midline on the bottom edge and the height on the bottom edge coincide), and the distance from the point on the bisector of the corner to both sides of the corner is equal.

(2) prove: CA = CB,

∴∠A=∠B, ∫o is the midpoint of AB, ∴ OA = OB.

∵DF⊥AC,DE⊥BC，

∴∠AMO=∠BNO=90，

* in △OMA and △ONB

,

∴△OMA≌△ONB(AAS)，

∴OM=ON.

(3) solution: OM=ON, OM ⊥ ON. The reason for this is the following：

Connect CO, then CO is the center line on the side of AB.

∫∠ACB = 90 degrees,

∴OC=

AB=OB，

CA = CB，

∴∠CAB=∠B=45,∠ 1=∠2=45，∠AOC=∠BOC=90，

∴∠2=∠B,∵BN⊥DE，

∴∠BND=90，

∵∠ b = 45，∴∠ 3 = 45，∴∠3=∠B，∴ dn = nb。

∠ ACB = 90，∴∠ NCM = 90。 ∴∠∵bn⊥de DNC = 90。

∴ Quadrilateral DMCN is a rectangle,

∴DN=MC,∴MC=NB，

∴△MOC≌△NOB(SAS)，

∴OM=ON,∠MOC=∠NOB，

∴∠MOC﹣∠CON=∠NOB﹣∠CON，

That is ∠ mon = ∠ BOC = 90,

∴OM⊥ON.