The reason for this is the following：

∫△ABD and △BCE are equilateral triangles,

∴∠DBE=∠ABC=60 -∠ABE，AB=BD，BC=BE。

In △ABC and △DBE,

AB=BD∠DBE=∠ABCBC=BE，

∴△ABC≌△DBE(SAS).

∴DE=AC.

AC = AF，

∴DE=AF.

In the same way, ef = ad can be obtained.

Quadrilateral ADEF is a parallelogram.

(2)∵ Quadrilateral ADEF is a parallelogram,

When daf = 90, the quadrilateral ADEF is a rectangle,

∴∠FAD=90。

∴∠bac=360-∠daf-∠dab-∠fac = 360-90-60-60 = 150。

Then when ∠ BAC = 150, the quadrilateral ADEF is a rectangle.