Find angle $\alpha$ formed in a triangle between two squares

Question

So, this is a problem from my friend @dare2solve over at Instagram, he posted this last year, I encountered it again today and decided to post it here. I have posted my own approach as an answer, please comment your own approaches as well, in particular if there's a way to use analytic geometry to solve it but any approach is welcome!

Answer 1

Because of the 45 degrees, BXDG is a straight line.

Note that BXDG is the perpendicular bisector of AC. Then, CG = GA = AC. That is, $\triangle ACG$ is equilateral with $\angle ACG = 60^0$. Therefore, $\angle DCG = 15^0$.

By symmetry, $\angle ECD = \angle DCG = 15^0$ and also $CE = CG= AC$

By now, we have all the tools to find $\angle AED = AEC + \angle CED = 75^0 + 30^0$

Answer 2

This is my own approach. I'll add the explanation below too.

Here's how I go about it:

Please note that I forgot to mark one of the points/vertices above the missing angle, therefore I'll be referring to it as point E throughout my explanation

1.) Label all the points appropriately as shown in my drawing, also label the point of at which the vertices of both squares touch at as point $D$. Now connect $A$ and $C$ via segment $AC$. Notice that $\angle CDA=\angle BDA=(360-90)/2=135$. This immediately tells us that $\triangle BDA$ and $\triangle CDA$ are congruent via the SAS property. It follows that segment $AB=BC=AC$. Thus proving that $\triangle ABC$ is equilateral as well as the fact that $\angle DAC=\angle DAB=30$ each.

2.) Connect point $E$ and point $C$ via the segment $CE$ and split the triangle $\triangle ABC$ into $\triangle ECB$ and $\triangle ECA$ respectively. Notice once again that $\angle EDC=\angle ADC=135$. This proves that $\triangle EDC$ and $\triangle ADC$ are also congruent via the SAS property, implying that segment $AB=BC=AC=CE$. This proves that point $C$ is the circumcenter of $\triangle BEA$.

3.) Notice that, as $\triangle EDC$ is congruent to $\triangle ADC$, $\angle AEC=30+45=75$. This then implies that that $\angle EAC=75$ as well, therefore $\angle ACE=\angle BCE=30$. Following this, its easy to see that $\alpha=\angle BEC+\angle DEC=75+30=105$