Rotate the sides of triangle t = (ABC) about their middles, by an angle (phi). Extending them and taking intersections, defines a new triangle s = (GHI), similar to t. As the angle (phi) varies, the triangle s takes various positions and the following is true:

(1) The circumcircles d1, d2, d3 of triangles GAI, HBG and HCI respectively, have their radical axis coinciding with the symmedians of triangle s. Hence their radical center coincides with the symmedian point P of s.

(2) The other than the vertices of GHI intersection points of the three circles d1, d2, d3, build a triangle u = RST, similar to the cosymmedian triangle of t.

The first statement was proved in the reference above. Having identified the symmedians with the radical axes, the point P (symmedian point of rotating triangle s) is the radical center of the three circles d1, d2, d3. Then PS*PG =PH*PT = > (PS/PT) = (PH/PG) = (PX/PV). This shows that ST is parallel to XV, which is a side of the cosymmedian triangle VWX of GHI. This completes the proof. Look at Cosymmedian.html for a picture of the cosymmedian triangle.

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