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.
 A: Build a parallelogram $ABDE$. Then $$\angle DEA = \angle ABD = 135^\circ = \angle DCA,$$ hence $ADEC$ is cyclic. Moreover, $\angle EAD = \angle BDA = x$, hence $$\angle CAE = \angle CAD - \angle EAD = 2x-x=x.$$ As a consequence, $CE=DE$ because the angles subtended by arcs $CE$, $DE$ of the red circle are equal.

Now, it is easy to calculate that angle $CBD$ equals $90^\circ + x$ and that the concave angle $CED$ equals $180^\circ + 2x$. Since $CE=DE$, it follows that $B$ lies on the circle with center $E$ and radius $CE=DE$. Hence $\angle DEB = 2\angle DCB = 90^\circ$. Since $DE=BE$, the right triangle $DEB$ is isosceles.
Let $F$ be the midpoint of $BE$. Let $G$ be the midpoint of $BD$ and $H$ be the midpoint of $BG$. Easy to see that triangles $GFB$ and $GFH$ are isosceles right triangles, hence $HF = HB = \frac 12BG = \frac 14 BD$, so $HD = BD - BH = 4HF - HF = 3HF$. This yields $\tan x = \frac{HF}{HD} = \frac 13$ so the answer is $$x = \arctan \frac 13.$$
A: I have a trigonometry solution. Looking at the form of the answer, I don't think there will exist a purely synthetic approach.
With some angle chasing, we get that
$$\angle ADC=45-2x$$
$$\angle ABC=45-x$$
Using law of sines on $\Delta ACD$, we get
$$\frac{AC}{\sin (45-2x)}=\frac{AD}{\sin 135}$$
Using law of sines on $\Delta ABD$, we get
$$\frac{AB}{\sin (x)}=\frac{AD}{\sin 135}$$
Combining these two equations gives
$$\frac{AC}{\sin (45-2x)}=\frac{AB}{\sin (x)}$$
$$\frac{AC}{AB}=\frac{\sin (45-2x)}{\sin (x)}$$
With simple trig definitions on right triangle $\Delta ABC$, we get that
$$\sin (45-x)=\frac{AC}{AB}$$
Hence,
$$\frac{\sin (45-2x)}{\sin (x)}=\sin (45-x)$$
Using angle sum/difference identities along with others, this simplifies to
$$\frac{\sqrt{2}}{2}(\cos (2x)-\sin (2x))=\frac{\sqrt{2}}{2}(\sin (x)\cos (x)-\sin^2 x)$$
$$\cos^2 (x)-\sin^2 (x)-2\sin (x)\cos (x)=\sin (x)\cos (x)-\sin^2 (x)$$
$$\cos^2 (x)=3\sin (x)\cos (x)$$
Note that if $\cos (x)=0$, then $x=\frac{\pi}{2}$. However, it is clear by the diagram that $x$ is acute (we can get even better bounds, but it is unnecessary).
Hence, we have $\cos (x)\neq 0\implies$
$$\cos (x)=3\sin (x)$$
$$x=\boxed{\cot^{-1} (3)}$$
A: 
Hints: You can easily see that:
$2x+\widehat {ADC}=45^o$
Now try to find another relation between these two angles using data the figures give.
A: 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, $2x \approx 37^\circ$
