# How is this angle measured in the triangle?

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?

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