### 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?

Explanation

[Honsberger, p. 33] attributes the following observation to Ian McGee, University of Waterloo:

 At points A and B on a circle, equal tangents AP and BQ are drawn as depicted in the applet. Then AB bisects PQ.

### 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?

Note that, because of the symmetry, the statement is obvious if AB is a diameter of the circle. This is quite surprising that it remains true when A and B are selected on the circle randomly.

Extend AP beyond A to R so that AR = AP. Let BQ meet AP in T, AB meet PQ in S. Now, tangents TA and TB from T to the circle are equal, and since AR = AP = BQ, we also have TR = TQ. Two isosceles triangles ABT and RQT share the same angle at the apex and are, therefore, similar. It follows that QR||AB, or QR||AS. In ΔPQR, A is the midpoint of side PR and AS is parallel to side QR, it is thus a midline of the triangle. Its other end S is then a midpoint of side PQ.

### References

1. R. Honsberger, In Pólya's Footsteps, MAA, 1997