## Symmedian property

Let b = (QRS) be the tangential triangle of a triangle a = (ABC). On the sides of b, in the same orientation and starting from the vertices of a, take segments AG, BF, etc. equal to a segment DE. Then, the radical axis of the circumcircles of triangles (ACG) and (ABF) coincides with the symmedian AQ of triangle ABC through A.

The following remarks lead to the proof.
(1) The centers I, J of the circumcircles c1 = {ACG}, c2 = {BFA} are, respectively, on the bisectors MR, MS of triangle b. The segments MI, MJ project orthogonally to AH, BL, respectively, which are half the length x of AG=BF=DE. Thus, triangle c = (JMI) remains similar to itself as x varies.
(2) The middle of segment IJ moves on line MA, as x varies. This, because J projects orthogonally to K, symmetric of L with respect to the bisector MS of the angle at S. Thus, AK = AH.
(3) The radical axis of c1, c2, passing through A and having a fixed direction, orthogonal to the direction of IJ, is the same line, independent of the magnitude of x.
(4) In the particular case, where x is the length of OP, triangle (JMI) coincides with triangle (PMO). Then AQ is the medial of OP and this identifies it with the line joining the contact point A, with the opposite vertex Q of triangle b. To see this, one can use the discussion on bisectors, exposed in TriangleBisectors.html .

The previous line, identified with AQ, is the symmedian of triangle ABC. This is discussed in the file Antiparallels.html .

For an application of the above discussion to a question of triangle-geometry, look at the file RotatingSides.html .