Angle chasing should lead to a solution. Here is some detailed computation of angles, well it may depend on the picture. First of all some notations. Let $x,y,z$ be the angles of the given $\Delta ABC$ in $A,B,C$. Yes, $x=y$, but we use this later only.
The given rotation is marked by a prime, so $A\to A'=A$, $B\to B'$, $C\to C'$.
Let $u$ be the used rotation angle, so it is the angle between any two segments, one obtained from the other one by rotation:
$$
u
=\sphericalangle(AB,A'B')
=\sphericalangle(AC,A'C')
=\sphericalangle(BC,B'C')\ .
$$
To write formulas in one breath, let $S$ be the the intersection $S=BC\cap B'C'$. We define $Q$ as the intersection $BB'\cap CC'$. Then:
$$
\begin{aligned}
\widehat{B'QC}
&=
\widehat{QBC}+\widehat{QCB}
=
\widehat{B'BS}+\widehat{SCC'}
\\
&=
(180^\circ -\widehat{B'SB}+\widehat{SB'B})
+
(180^\circ -\widehat{C'SC}+\widehat{SC'C})
\\
&=
180^\circ -\widehat{SB'B}-\widehat{SC'C}
\\
&=
180^\circ
-\left(y+\left(90^\circ-\frac u2\right)\right)
-\left(z-\left(90^\circ-\frac u2\right)\right)
=
180^\circ-y-z
\\
&=x=\widehat{CAB}=\widehat{B'AC'}\ .
\end{aligned}
$$
In particular, $CQBA$ is cyclic,
and $AC'B'Q$ cyclic,
so $Q$ is on the circles $\odot(ABC)$ and $\odot(A'B'C')$.
So far $x,y$ are not constrained to equality.
In the second part we need $x=y$, and it is simplest to use complex numbers and compute. We place the origin $O$ in $O=A$, and let the $Ox$ axis to be the line $AB$. The lengths of the sides in $\Delta ABC$ are by the sine theorem proportional to $AC:BC:AB=\sin x:\sin x:\sin 2x$, so we are placing $B$ in $\sin 2x\in\Bbb C$. Then we can compute all affixes, but let us do so only for the needed points. Since $v:=u/2$ will appear, we try to write all formulas by using $v$ instead of $u$.
$$
\begin{aligned}
B &= \sin 2x\ ,\\
B' &= \sin 2 x\cdot e^{i\; 2v}\ ,\\
P &=\frac 12(B+B')=\sin 2x\cos v\cdot e^{i\; v}\ ,\\
C &= \sin x\cdot e^{ix}\ ,\\
C' &= \sin x\cdot e^{i(x+2v)}\ ,\\
D &= |AD|\cdot e^{i(x-\pi/2)}=-i\cos x\cdot e^{i\;x}
\\[3mm]
\frac{C'P}{DP}
&=
\frac
{C'-P}
{D-P}
\\
&=
\frac
{\phantom{(-i)}\sin x\cdot e^{i\; v}\cdot ( e^{i(x+v)} - 2\cos x\cos v)}
{(-i)\cos x\cdot e^{i\; v}\cdot ( e^{i\; (x-v)} - 2i\sin x \cos v)}
\\
&=
\frac
{\phantom{(-i)}\sin x\cdot e^{i\; v}\cdot ( -\cos(x-v) + i\sin (x+v) }
{(-i)\cos x\cdot e^{i\; v}\cdot ( \cos(x-v) -i\sin(x+v) }
\\
&=-i\tan x\ .
%
\end{aligned}
$$
So the point $C'$ is obtained from $D$ by a rotation of angle $-\pi/2$ around $P$, composed with a $\tan x$-rescaling. (This is what we have expected, since $C'PQD$ cyclic is equivalent to $x=\widehat {PQC'}=\widehat{PDC'}$.)
$\square$
Note: Usually, in similar situations, the algebra involving complex numbers has a point / bridge of cancellation, that suggests which geometric construction has the "same bridge". In this case, the algebra is straightforward. But the geometry is not so simple. For instance, the angle $x+v=\widehat{C'AB'}+\widehat{B'AP}=\widehat{C'ABP}$ is an angle in the interesting triangle $PAC'$. It can be used to compute $PC'$ (squared) using Pythagoras, thus avoiding the "bad" information on the other two angles. But this computation is more or less contained in the above. I have an argument obtained by "reversing the order", constructing $C'$ so that an angle $x$ in $Q$ is ensured, and showing that it is "our point", but this has also esthetic impediments.
