<s>[INST] Calculate angle $x$ in the figure For reference:

My progress:

\begin{align*} 
& AH \perp FD \\
& \triangle AFD \text{ is isosceles} 
\quad 
\therefore \measuredangle BFA  = \measuredangle FDA = x \\
& AF = FD \\ 
& \measuredangle HBA = 180-135^\circ = 45^\circ 
\quad 
\therefore \triangle HBA \text{ is isosceles}
\end{align*}
I drew some auxiliary lines but it wasn't enough to reach the solution.
 [/INST] Once again, Pythagoras theorem does the job.

Say, $BC = a, AC = b$. Drop a perp from $B$ to $AD$. Notice that right triangles $\triangle ABC \cong \triangle ABH$, given $\angle ABC = \angle BAH = 45^\circ - x$ and $AB~$ is hypotenuse to both triangles.
So, $AH = BC = a, BH = AC = b$
Similarly notice that $\triangle DBE \cong \triangle BDH$. So, $DE = BH = b, DH = BE$ but $BE = CF = DF = a + b$
That leads to $DF = a + b, AF = a + 2b, AD = 2a + b$
Applying Pythagoras,
$(a+b)^2 + (a+2b)^2 = (2a+b)^2 \implies (a + b) (a-2b) = 0$
So we get $a = 2b$ and that shows $\triangle ADF$ is $3:4:5$ triangle.
Hence, $2x \approx 37^\circ$
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