The intersection points of an ellipse with a line parallel to the line which contains the two foci of the ellipse, and the foci are concyclic.

Question

I was playing around with ellipses in a graphing app and I found this property which seems to be true, but I can't find a way to prove it.

Let A and B be the foci of the ellipse, draw a line parallel to AB that intersects the ellipse at points C and D. It seems that A,B,C and D are concyclic.

I've tried using analytic geometry but the equations get too stuffy very fast.

Answer 1

I haven't tried it, but you could find the circle generated by A, B and C in the plane and then prove it intersects the ellipse in such a way that the intersection of the circle with the ellipse, D, is parallel to AB. If you prove for a nice ellipse where AB is along the x axis and the centre of the ellipse is at the origin, then all you have to do is show D is a reflection of C in the Y axis. The result will hold generally after scaling, rotations and translations.

Answer 2

In an isosceles trapezoid, the two base edges have the same perpendicular bisector. The perpendicular bisector is a line of reflective symmetry of the trapezoid.

Now take one of the other edges and find the intersection of its perpendicular bisector with that line of symmetry. Reflecting the figure through the line of symmetry, the perpendicular bisector of the fourth edge intersects the line of symmetry in the same point.

In other words, the perpendicular bisectors of all four sides meet at a single point. That point is the center of the circle that passes through all four vertices.