I read "How to solve it" by George Polya. I do not understand how he gets to certain solutions. Like the chapetr of "Auxillary Solution". Given the picture below:

Leave the angle at the top $ A $ to be called $ alpha $. Polya says looking at the diagram and the isosceles triangles $ triangle {ACE} $ and $ triangle {ABD} $ we can maintain that $ angle {DAE} = frac { alpha} {2} + 90 ^ circ $.

I just do not understand how this is possible. All I can understand is that $ angle {CAE} = angle {CEA} $ and $ angle {BAD} = angle {BDA} $since they are both isosceles. But where will I get $ 90 ^ circ $ of?

Do I miss an important theorem in geometry?