Euclidean geometry, Prove that either $X, Y, Z, T$ belong to the same circle or $X, Y, Z, T$ belong to the same line. 
Let the quadrilateral $ABCD$ have pairs of non-parallel opposite
sides. $O$ is the intersection of $AC$ and $BD$. The circumcircle of
the triangles $OAB$, $OCD$ intersect at $X$ different from $O$. The
circles The circumcircle of the triangles $OAD$, $OCB$ intersect each
other at $Y$ different from $O$. The circles with diameters $AC$, $BD$
intersect at $Z$, $T$. Prove that either $X, Y, Z, T$ belong to the
same circle or $X, Y, Z, T$ belong to the same line. 

This is an assignment from my maths teacher, when i have learned about Harmonic Division.
I can't solve this geometry problem. Any hints are appreciated.
 A: Invert the configuration around $O$ with arbitrary radius, and let $P'$ be the image of the point $P$. The position of the images I will introduce now can be proven with well-known properties of inversion that I will outline, and you might want to look at the diagram below in order to visualize the resulting construction after the inversion.
Notice that $(\triangle OAB), (\triangle OCD), (\triangle OAD), (\triangle OBC)$ will be sent to lines $A'B', C'D', A'D', B'C'$ respectively since the four circumcircles contain the inversion center $O$. Thus, $X'=A'B'\cap C'D'$ and $Y'=D'A'\cap B'C'$. At the same time, the circles $\Gamma_1, \Gamma_2$ with diameters $AC, BD$ respectively will be sent to the circles $\Gamma_1', \Gamma_2'$ with diameters $A'C', B'D'$ since their centers lie on $AC\ni O, BD\ni O$ respectively. Finally, $X,Y,Z,T$ will lie either on a circle or on a line if and only if $X',Y',Z', T'$ do so.

Thus, the equivalent problem reads:

Given a quadrilateral $A'B'C'D'$, define $X'=A'B'\cap C'D'$ and $Y'=D'A'\cap B'C'$. Also, let $T',Z'$ be the intersections of the circles with diameters $A'C'$ and $B'D'$. Prove that $X',Y',Z',T'$ are either concyclic or collinear.

Yet this follows from Gauss-Bodenmiller Theorem.

 For the proof: We will prove that the desired circle has the diameter $X'Y'$. Consider now the orhocenters $H_1,H_2, H_3, H_4$ of triangles $X'A'D', X'B'C', Y'B'A', Y'C'D'$ respectively. The idea is to take any orthocenter and show that it has the same power with respect to all three circles. Hence all four orthocenters lie on all the radical axes. This implies the conclusion.

