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Given a parallelogram ABCD and points M on AB and N on CD. K is a point on MN. The circumcircles of triangles AMK and CNK, if not tangent, meet in point P different from K. Interestingly, P always lies on the diagonal AC. This is a generalization of a similar property established for a square. The proof requires no change even in the case of the parallelogram.

However, there is no reason to stop investigating the diagram depicted by the applet. The applet is suggestive of additional properties.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run?

A simple observation is that the angle formed by the line KP with the diagonal AC is independent of K (assuming of course all other points fixed.) This follows from ∠APK + ∠AMK = 180° and the fact that ∠AMK is fixed.

The latter property has an engaging implication. The centers of the two circles lie on the perpendicular bisector of the common chord KP, so that the line of the centers remains parallel to itself as K slides along MN.

More than that, as K slides along MN, the center line also preserves the length. This is because the centers move along fixed perpendicular bisectors of segments AM and CN.

In the limit, when the two circles become tangent and the points P and K coincide, the common tangent two the two circles through P (or, which is the same, K) maintains the same angle with the diagonal as was formed by the non-degenerate chord KP. This angle only depends on the parallelogram ABCD and the angles formed by MN with the sides of the parallelogram, but not on the location of the point P (= K) on the diagonal AC. So that if MN moves to a different location staying parallel to itself so does the common tangent of the two circles whenever P = K.