The circumcenter Q of the tangential triangle DEF of triangle ABC is on the Euler line of ABC. Hence the homothety center of the orthic KLM and the tangential DEF of ABC (X(25) according to Kimberling) is also on the Euler line of ABC.
- Draw from the vertices of the tangential triangle parallels to the sides of ABC to build GIJ homothetic to ABC. Since BCF, CDA, AEB are isosceli the lines joining D, E, F with the circumcenter O of ABC are orthogonal to corresponding sides of GIJ and easily identified as its altitudes. Hence O is the orthocenter of GIJ.
- It follows that the circumcircle (c) of the tangential DEF is the Euler circle of GIJ, hence the circumcenter Q of DEF and O are on the Euler line of GIJ.
- The Euler lines of ABC and GIJ are parallel and have O in common, hence they coincide. Remark: The figure above shows also the contact point of the Euler circle of DEF with the circumcircle of ABC. This is the focus of the Kiepert parabola (X(110) in the enumeration of Kimberling). Notice also that the homothety ratio/center of the orthic to the tangential is the same with the homothety ratio/center of ABC to GIJ.
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Tangential circumcircle ![[0_0]](TangentialCircumcircle_files/TangentialCircumcircle_0_0_0.png)
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