Geometry question on angle chasing concerning 3 squares

Question

I am kind of stuck on this problem. We know that the 9 points present in this sketch form three squares, HBAI, CFGB and DEFC. We also know, that the lines DI and AF intersect in S. The question for this problem is to find the angle ASD.

My approach for this problem: Because CF is common for the two squares, we know the sides of the squares CFGB and DEFC are the same. Then I can rewrite the angle as: 180 - sin^-1 (CF/AC) - sin^-1 (AI/AD). Then I can rewrite this using Pythagoras but I stay stuck. Assuming from the sketch I can see that BG is 4/3 of AI, and then I come easily at 135 degrees for the angle. Unfortunately I am unable to prove it, and I can’t know whether my assumption is right.

Answer 1

Let us assume $\angle{AID}=x$ and $\angle{AFC}=y$. It is clear that $\angle{ASD} = x+y$. Also let us consider $b$ as length of smaller square side and $a$ as side length of bigger squares. We have

$\tan{x} = \frac{2a+b}{b} = 1 + \frac{2a}{b}$.

$\tan{y} = \frac{a+b}{a} = 1 + \frac{b}{a}$.

Now, $\tan{\angle{ASD}} =\tan{(x+y)}=\frac{\tan{x}+\tan{y}}{1-\tan{x}\tan{y}} = \frac{2+\frac{2a}{b}+\frac{b}{a}}{1-(1 + \frac{2a}{b})(1 + \frac{b}{a})}=\frac{2+\frac{2a}{b}+\frac{b}{a}}{1-(1 +\frac{2a}{b}+\frac{b}{a} +2)}=-1$.

So the angle is 135 degrees.

Answer 2

By this construction, we know $AF$ passes through $M$ by side ratios on triangle $AFC$ to $M$ equaling $a+b\over b$. By green congruency we can infer $AMN$ is a $45,45,90$ triangle. And $ID\parallel AN$ which gives $\angle MHD=45^\circ$.