As shown in figure 1, rotate Δ δBEC counterclockwise by 90 degrees around point C to obtain Δ B 'e 'c, which is connected with DE'

Then B'E'DC is a rectangle, CE=CE', ∠ECE'=90 degrees, ∠CEE'=45 degrees.

And BE=AC, so AE=CB'=DE', AE∨DE',

So the quadrilateral AEE is a parallelogram, AD∨EE'

So ∠AOE=∠OEE'=45 degrees. ?

Another solution (Figure 2)

Let D'E be perpendicular to AB, and D'E=BE pass through point e.

Link AD', DD'

The triangle ADD' that is easy to prove is an isosceles right triangle, and CED'D is a parallelogram.

∠AOE =∠ addition' =45 degrees.

The idea is to make an angle of 45 degrees, that is, to make an isosceles right triangle.