It is proved that the intersection point e makes EH vertical AC intersect with AC at h, and connects BD with AC at o,

In a square ABCD, AC⊥BD, AC=BD, OB= half BD= half AC,

Similarly, the quadrilateral AEFC is a diamond,

∴AC=CF,AC∥EF，

∵EH⊥AC，

∴∠BOH=∠OHE=∠OBE=90，

∴ Quadrilateral Beho is a rectangle,

∴EH=OB，

Eh = half AC= half AE,

In the right triangle AHE,

So ∠ EAH = 30, (in a right triangle, the opposite side of the 30-degree angle is half of the hypotenuse)

That is, ∠ EAB = ∠ cab-∠ EAH = 45-30 = 15.

So the answer is 15.

Please give good comments.