For the centroid, the circumcenter and the orthocenter it is quite clear that if the lines intersect on BD for one position of C, then the same is true for all other positions as well. (What is lost or gained on the left from the common point of intersection, is gained or lost on its right.)

I've been thinking about it, but I don't see this at all. Can anyone help?

Let's take the centroids. Assume the feet of the medians from A are M (for triangle ABC) and N (for triangle ACD). Then we think of N and M as functions of C. M is the midpoint of BC, N is the midpoint of CD. When C moves by D, both M and N move by D/2. The centroids are 1/3 up from M and N on AM and AN, respectively. Their speeds are 2/3 of those of M and N, which is D/3 in both cases. In addition the motion of both centroids is parallel to BD.

Now consider lines through the centroids parallel to AD and AB. Each of these lines moves along with the corresponding centroid at the same rate and in the same direction. This is a sort of dynamic parallel transform. This implies that the point of intersection of those two lines also moves at the same velocity: 2/3 the speed of C and in the same (plane) direction.