Similar triangles constructed by circumcircles of triangles in another triangle $ABC$ I have trouble solving this romanian olympiad question. I translated it:

Let a point $D$ be chosen inside a triangle $A B C$. The centers
of the circumcircles of the triangles $A B D, B C D$ and $CAD$ may be $C^{\prime}, A^{\prime}$ and $B^{\prime}$ respectively.
Let the straight line $AD$ intersect the circumcircle of $BCD$ in $A^{\prime \prime}$. Let $B^{\prime \prime}$ and $C^{\prime \prime}$ be constructed accordingly.
Prove that the triangles $A B C^{\prime \prime}, B C A^{\prime \prime}, C A B^{\prime \prime}$ and $A^{\prime} B^{\prime} C^{\prime}$ are similar.
Is it possible, for any triangle $A B C$ to choose the point $D$ in such a way
so that these four triangles are also similar to the triangle $A B C$?

Here is a visualization I constructed with geogebra

I started with the observation, that the triangles in the circumcircles coincide with one side of $ABC$. And if the triangle $ABC$ is equilateral and $D$ is exactly where the bisectors of its angles meet, the triangles have all the same size. My only trouble is proving it and creating a relation between $ABC$ and the triangle constructed by the midpoints of the circumcircles.
 A: 
As $BDAC''$ is concyclic, $\angle C''AB = \angle C''DB$, which is equal to $\angle CDB''$ (since lines $CDC''$ and $BDB''$ are straight). And as $CDAB''$ is concyclic, $\angle CDB'' = \angle CAB''$. Also, as $BDCA''$ is concyclic, $\angle C''DB = \angle CA''B$. Proceed similarly with other angles and you will get that $\triangle A''BC \sim \triangle AB''C \sim \triangle ABC''$.
For $\triangle A'B'C'$, observe that $C'A'$ is the perpendicular bisector of the line segment $BD$ (since $C'A'$ is the line joining the centres of those two circles), and similarly $B'A'$ is the perpendicular bisector of $CD$. Therefore, in quadrilateral $A'EDF$, $\angle EDF + \angle EA'F = 180^\circ$. And also $\angle BDC + \angle BA''C = 180^\circ$. From these two equations we get $\angle C'A'B' = \angle BA''C$. Proceeding similarly with the other angles, we can prove the similarity of $\triangle A'B'C'$ as well.
Now think about the choice of point $D$ if I want $\triangle ABC$ similar too. Comment after that.
