I was looking at this web page
http://www.cut-the-knot.com/ctk/CompleteQuadrilateral.shtml
and it states, near the bottom:

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?

It is simple but to a different degree for the centroids, orthocenters and the circumcenters.

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.

Ok, thank you, I see that now. Basically, the two lines through C and C itself move in tandem. I was getting confused becuase sometimes, when I clicked on the java applet to move the point C, point A would move as well, which completely obscures the argument.

I see the argument for the centroids, that's pretty clear. I see how the argument goes for the orthocentres as well. Do you have a nice one for the circumcentres? (Other than appealing to the Euler lines of triangles ABC and ACD.)

>Do you have a nice one for the circumcentres?

Perpendicular bisectors of the bases move with equal velocities. For each of the triangles ABD and ACD, this movement causes the circumcenters to glide along the perpendicular bisectors of AB and AD, which do not move. These motions, too, have fixed velocities. Their components parallel to the base are equal. Only these components are responsible to the point of intersection of the lines through the circumcenters with the base. Thus the two lines intersect on the base for all positions of C, if they do for one. This one can be taken as the foot of the altitude from A, in which case they meet in the midpoint of BD.

Makes sense, thank you. Clever argument!