Fact. If $x$, $y$, $z$ are points on the complex plane, then the circumcenter of $\triangle xyz$ is the point
$$ i \frac{x(z\overline{z} - y\overline{y})+ y(x\overline{x}-z\overline{z})+z(y\overline{y}-x\overline{x})}{x(\overline{z} - \overline{y})+ y(\overline{x}-\overline{z})+z(\overline{y}-\overline{x})}$$
where "$\overline{x}$" indicates the complex conjugate of $x$.
Let us consider $\triangle abc$, defining $d$, $e$, $f$ on sides $bc$, $ca$, $ab$, respectively:
$$d := b + \alpha(c-b) \qquad e := c+\beta(a-c) \qquad f := a+\gamma(b-a)$$
for real scalars $\alpha, \beta, \gamma$.
Let $p$, $q$, $r$ be the respective circumcenters of $\triangle aef$, $\triangle dbf$, $\triangle dec$; for simplicity, we take $a$, $b$, $c$ on the unit circle so that $\overline{a}=1/a$, etc. Then,
$$\begin{align}
p &= i\;\frac{c(b-a)+ \beta b ( a-c) + \gamma c( a - b )}{b-c} \\[6pt]
q &= i\;\frac{a(c-b)+ \gamma c ( b-a) + \alpha a( b - c )}{c-a} \\[6pt]
r &= i\;\frac{b(a-c)+ \alpha a ( c-b) + \beta b( c - a )}{a-b}
\end{align}$$
We deduce
$$\frac{p-q}{a-b} = \frac{q-r}{b-c} = \frac{r-p}{c-a} =: u$$
which, by taking the modulus, proves $\triangle abc \sim \triangle pqr$. A little algebra reveals that
$$u + \overline{u} = 1$$
Now ... A point, $m$, (which may or may not be the Miquel point) is the center of spiral similarity from $\triangle abc$ and $\triangle pqr$, if and only if
$$\frac{a-m}{p-m} = \frac{b-m}{q-m} = \frac{c-m}{r-m}$$
(This characterization is what inspired me to approach the problem in the complex plane.) Solving for $m$ via the first equality gives
$$m = \frac{aq-bp}{(a-b)-(p-q)} = \frac{r - c u}{1-u}$$
whence
$$m-r = \frac{u(r-c)}{1-u} = \frac{u}{\overline{u}}(r-c) \qquad\qquad
\overline{m}-\overline{r} = \frac{\overline{u}}{u}(\overline{r}-\overline{c})$$
so that
$$|m-r|^2 = (m-r)(\overline{m}-\overline{r}) = (r-c)(\overline{r}-\overline{c}) = |c-r|^2$$
indicating that $c$ and $m$ are equidistant from $r$: thus, $m$ is on the circumcircle of $\triangle dec$. By symmetry, it is also on the circumcircles of $\triangle aef$ and $\triangle dbf$; the center of spiral similarity is in fact the Miquel point.
First, since $CD \perp AB$ and $DK$ and $DN$ are angle bisectors to the right angles
$\angle \, ADC$ and $\angle \, BDC$, then $$\angle \, KDN = \angle \, KDC + \angle \, NDC = 45^{\circ} + 45^{\circ} = 90^{\circ}$$
However, $\angle \, KCN = 90^{\circ}$ so the quadrilateral $CKDN$ is inscribed in a circle.
Next, prove that $KL\, || \, CB$ and $NL\, || \, CA$ using the properties of angle bisectors and the similarity between triangles $ABC, ACD$ and $BCD$. Indeed, since $DK$ is a bisector of the angle at vertex $D$ of triangle $\Delta \, ADC$, we apply the theorem that
$$\frac{AK}{KC} = \frac{AD}{DC}$$ But triangles $\Delta \, ACD$ is similar to $\Delta \, ABC$ so $$\frac{AD}{DC} = \frac{AC}{CB}$$ so
$$\frac{AK}{KC} = \frac{AC}{CB}$$ By the fact that $CL$ is an angle bisector of the angle at vertex $C$ of triangle $\Delta\, ABC$
we have that
$$\frac{AC}{CB} = \frac{AL}{LB}$$ so consequently
$$\frac{AK}{KC} = \frac{AL}{LB}$$
which by Thales' intercept theorem implies that $KL \, || \, CB$. Analogously, one can show that $NL \, || \, CA$.
Then quad $CKLN$ is a rectangle, so $KL =NL$ as diagonals in a rectangle. Therefore the point $L$ also lies on the circumcircle of quad $CKDN$ and $KN$ and $CL$ are diameters of the said circle.
Best Answer
Let $AB$ be transformed into $AB'$ by a rotation and scale factor. Then $AC$ is transformed into $AC'$ by the same rotation and scale factor.
Therefore $ABB'$ and $ACC'$ are similar.
Then angles $ABD$ and $ACD$ sum to $\pi$. Therefore $ABCD$ is a cyclic quadrilateral as required.