Let be the centroids of , respectively. Note that as , we have . Since is a cyclic quadrilateral, it must be isosceles, so . As and concur at the midpoint of , it must hold true that .

Now from either the Appolonius Theorem for Medians or some Law of Cosines (if you're one of those people like me who keeps forgetting the formula), it is true that

\[\begin{align*}\dfrac{2AB^2+2BD^2-AD^2}4=BG^2&=CG^2=\dfrac{2AC^2+2CD^2-AD^2}4\\\implies AB^2+BD^2&=AC^2+CD^2.\end{align*}\]

It is given that , so through some manipulation . Thus the roots of and concur, whence . However, if and , then , contradiction. Therefore as desired.

Let be the circumcenter of and its orthocenter. Furthermore, let be the midpoint of and the foot of the altitude from to . Note that from some simple angle chasing we get , so considering the spiral similarity sending the former triangle to the latter gives that . As is the perpendicular bisector of , , so , the orthocenter of , is the reflection of across . Thus, quadrilaterals and are similar. Now, let be the distance from to the line , and note that


Thus the distance is fixed, and so lies on a line parallel to . All that remains is to show that lies on this line, but this is easy, as it is well known that and that the distance from to is , where is the circumradius of .