How is this angle measured in the triangle?

Question

I am reading George Polya's "How to Solve It". I cannot understand how he is getting to certain solutions. Like the one from the chapter of "Auxiliary Solution". Given the image below:

Let the angle at vertex $A$ be called $\alpha$. Polya says by looking at the diagram and the isoceles triangles $\triangle{ACE}$ and $\triangle{ABD}$ we can deduce that $\angle{DAE} = \frac{\alpha}{2} + 90°$.

I just don't understand how that is possible. All I can figure out is that $\angle{CAE} = \angle{CEA}$ and $\angle{BAD} = \angle{BDA}$, since they are both isoceles. But where will I get $90°$ from?

Am I missing some important theorem in geometry?

Answer 1

Exterior angle of triangle = sum of 2 interior opposite angles $$ \angle ACB = 2 \angle EAC \\ \angle ABC = 2 \angle DAB \\ $$ $$ \Rightarrow \angle EAC + \angle DAB = 0.5(\angle ACB + \angle ABC) = 0.5(180-\alpha) $$ $$ \angle DAE = 90-0.5\alpha + \alpha $$

Answer 2

$$\angle DAE = \angle EAC + \angle DAB + \alpha$$ $$= \frac{180°-\angle ACE}{2}+\frac{180°-\angle ABD}{2}+\alpha$$ $$=\frac{\angle ACB+\angle ABC}{2}+\alpha$$ $$=\frac{180°-\alpha}{2}+\alpha$$ $$=90°+\frac{\alpha}{2}$$