## Diameters and Chords

The applet below illustrates an ancillary problem that is useful for solving a curious problem discussed elsewhere. I preserve most of the notations to make the references easier.

 Let O be the center of a circle with diameters BBt, CCt and MtNt and chords BAb and CAc. Assume that BAb intersects MtNt in M and CAc intersects MtNt in N. Kb is the second point of intersection of NBt with the circle. Kc is the second point of intersection of MCt with the circle. Prove that Ab and Ac coincide if and only if so do Kb and Kc.

### 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?

Solution ### 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?

This is plainly a variant of Pascal's Hexagram Theorem.

Since A's and K's are interchangeable, we only have to prove that if Ab = Ac = A, say, then, too, Kb = Kc.

Consider the hexagon ABBtKbCtC. N is the intersection of sides AC and BtKb, O is the intersection of sides BBt and CtC. Let M' be the intersection of AB and KbCt. According to Pascal's Theorem, the three points N, O, and M' are collinear so that M' is the intersection of AB and MtNt. Therefore M = M'. The fact that, by the construction of Kc, M is also the intersection of MtNt with KcCt, implies that KcCt and KbCt are one and the same line so that, too, Kb and Kc are one and the same point.

As we mentioned at the outset, this result has repercussions to a theorem about a line passing through the circumcenter of a triangle. ### Pascal and Brianchon Theorems 