(1) Because the diagonals of the rhombus bisect each other vertically, they are divided into four identical parts.

AE=BO=DO

AE⊥CD again

∠ doc =∠ aed = 90。

Triangle docking congruent triangles AEC

So AC=DC=AD

So the triangular ADC is an equilateral triangle.

∠CAE=∠CDO=∠ADC/2=30

(2)

The vertices e and f of AEF of equilateral triangular paperboard fall on both sides of rhombic paperboard.

Explain that the condition of (1) is just met, that is, the triangle ABC and ADC are equilateral triangles.

So CDO = 2 * 60 = 120.

(3) Because AE⊥BC,AF⊥CD, Diaojiaolou E and F are the midpoint of BC and CD respectively.

Then AB=AC=AD

At the same time, AF and AE divide ∠DAC and ∠BAC equally.

Plus the four sides of the diamond are equal.

So it is still a triangle. ABC and ADC are equilateral triangles.

So ∠ EAF = (∠ DAC+∠ BAC)/2 = (60+60)/2 = 60.