(1) In a triangle, the sides of equal sides are equal to the angles, and then we can know that ∠ACB= 1/2∠ABC, and in a right triangle, two acute angles can be solved by complementation.

(2) A triangle with two equal sides is an isosceles triangle, which connects DB. According to the nature of isosceles trapezoid, the relationship between line segments and parallelism, it can be proved that AC = AF.

Solution: (1)∵AD∨BC,

∴∠DAC=∠ACB.

∫AD = DC，

∴∠DCA=∠DAC.

∴∠DCA=∠ACB= 1/2∠DCB。

∫DC = AB，

∴∠DCB=∠ABC.

∴∠ACB= 1/2∠ABC。

In △ACB, ∵AC⊥AB,

∴∠CAB=90。

∴∠ACB+∠ABC=90。

∴ 1/2∠ABC+∠ABC=90。

∴∠ABC=60。

(2) connect DB,

∫ in trapezoidal ABCD, AB=DC,

∴AC=DB.

In quadrilateral DBFA, DA∑BF, DA=DC=BF,

Quadrilateral DBFA is a parallelogram.

∴DB=AF，

∴AC=AF.

That is, △ACF is an isosceles triangle.