a) Let $O$ be the circumcenter of $ABCD$. It is well known that the midpoints of the sides of a quadrilateral $ABCD$ are the vertices of a parallelogram, hence the line segments joining the midpoints of two opposite sides have the same midpoint, $G$. The perpendicular lines from $O$ to $AB$ and $CD$ pass through the midpoints of these sides, therefore the perpendiculars dropped from $O$ and from the midpoints of two opposite sides onto their opposite side form a parallelogram whose center is $G$. It follows that the two perpendicular lines dropped from the midpoints of two opposite sides onto their opposite side intersect at the reflection of $O$ in $G$. The other two perpendiculars intersect at the same point.

b) We assume the angle $AXB$ to be acute, the other case being similar. The quadrilaterals $ABA'B'$, $CDC'D'$ and $ABCD$ being cyclic, we have $\angle XDC' \equiv \angle XDC \equiv \angle XAB \equiv \angle XA'B'$, hence $A'B'C'D'$ is cyclic.

c) Let $H_1$ and $H_2$ be the orthocenters of triangles $XAB$, and $XCD$, respectively. If $O''$ is the midpoint of $[H_1H_2]$, as $O''$ belongs to the midsegment of the trapezoid $H_1B'H_2D'$, $O''$ belongs to the perpendicular bisector of the line segment $[B'D']$. Similarly, $O''$ belongs to the midsegment of the trapezoid $A'H_1C'H_2$, hence to the perpendicular bisector of $[A'C']$. As $A'C'$ and $B'D'$ are not parallel, it follows that $O''$ is precisely the circumcenter of $A'B'C'D'$, i.e. $O''$ coincides with $O'$.

d) We have $\angle A'B'X \equiv \angle ABX \equiv \angle DCX$, hence $A'B' \parallel CD$. If $N$ is the midpoint of $[AB]$, then $NA' = NB'$, hence $N$ belongs to the perpendicular bisector of $[A'B']$. But so does $O'$, therefore it follows that $NO' \perp CD$. Similarly, $O'$ belongs to the perpendicular dropped from the midpoint of $[CD]$ on $AB$, hence $O'$ is the Mathot point of the quadrilateral.