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Explanation

### Two Circles in a Square

The applet suggests the following statement [Greitzer, p. 57, Honsberger, p. 118]:

 Assume points M and N are selected on the sides AD and BC of the square ABCD. Let K be an arbitrary point on MN. Besides K, the circumcircles of triangles AMK and CNK intersect at a point P. Prove that P always lies on the diagonal AC.

### 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.

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### Proof

Assume that M is strictly inside segment AD, while N is strictly inside segment BC. Assume the quadrilateral AMKP is convex. (Slight modifications may be necessary if those conditions do not hold.)

First, ∠APK + ∠AMK = 180°. Also ∠KNC = ∠KPC and ∠AMK = ∠KNC. Therefore, ∠APK + ∠KPC = 180°, which exactly means that P lies on the diagonal AC.

### Remark 1

Nathan Bowler has observed that the proposition holds not only for squares, but also for rectangles and, in fact, parallelograms.

### Remark 2

As an afterthought of handling a different problem it became clear that the diagram depicted by the applet offers more properties than has been suggested by either of the references. These will be investigated along with the extended problem.

### References

1. S. Greitzer, Arbelos, v 5, MAA, 1991
2. R. Honsberger, Mathematical Chestnuts From Around the World, MAA, 2001