### Pinning Butterfly on Radical Axes

The applet below illustrates a solution to the traditional Butterfly Problem that was communicated to me by Giles Gardam, a member of Australia's 2007 and 2008 IMO teams, and is due to his mentor, Ivan Guo, who won a gold medal at IMO 2004.

In a circle T, two chords AB and CD intersect at the midpoint of a third chord PQ. AD meets PQ in X, BC meets PQ in Y. The task is to show that M is also the midpoint of XY.

### Among the many browsers currently available, the Firefox is the one that - with sufficient permissions - still runs Java.

 What if applet does not run?

Reflect B and C and the circle T in M (by reflect in a point I mean a dilation by factor -1 about that point). B', C', and T' be the reflections of B, C and circle T, respectively. Note that as M is the midpoint of PQ, T' passes through P and Q.

By power of point M with respect to T, AM×BM=CM×DM. Thus AM×B'M=C'M×DM, so by converse of power of a point, AB'C'D is a cyclic quadrilateral. Let its circumcircle be T''.

We now consider the three radical axes of the three circles, which are PQ, AD and B'C', and concur by the radical axes theorem. Thus B'C' intersects PQ at X. The reflections of BC and PQ are B'C' and PQ, so considering their intersections, we have that X is the reflection of Y, thus M is the midpoint of XY. ### Butterfly Theorem and Variants  