Interesting geometry problems- By Mind Your Concept.

Let the large blue square be $ABCD$ with side length $1$ , the small red square be $AEFG$ , $\angle ADG = \alpha$ and the yellow angle $BFC=\theta$ . We note that $\triangle ABE$ and $\triangle ADG$ are congruent. Then $BF = BE-EF = \cos \alpha - \sin \alpha$ . By sine rule , we have:

$\begin{aligned} \frac {\sin \angle BFC}{BC} & = \frac {\sin \angle BCE}{BF} \\ \frac {\sin \theta}1 & = \frac {\sin (180^\circ - (90^\circ + \alpha) - \theta)}{\cos \alpha - \sin \alpha} \\ \sin \theta (\cos \alpha - \sin \alpha) & = \sin (90^\circ - \alpha - \theta) = \cos (\theta + \alpha) \\ \sin \theta \cos \alpha - \sin \theta \sin \alpha & = \cos \theta \cos \alpha - \sin \theta \sin \alpha \\ \sin \theta \cos \alpha & = \cos \theta \cos \alpha \\ \tan \theta & = 1 \\ \implies \theta & = \boxed{45^\circ} \end{aligned}$