## On the Euler-Circle theorem

The Euler circle is the circle passing through the middles A'', B'', C'' of the sides of triangle ABC. It passes also through: (a) the feet A*, B*, C* of the altitudes on the opposite sides, (b) the middles A''', B''', C''' of the segments joining the orthocenter H to the vertices of the triangle.

Consider the radii from the center O, of its circumcircle, to the vertices of the triangle ABC. Let A'B'C' be the triangle of the centers of the circles with diameters OA, OB, OC. The triangle A'B'C' has sides parallel and half the length of the corresponding sides of ABC. The altitudes of A'B'C' cut each other at the center M of the Euler circle, passing through A'', B'', C''. This because e.g. the diacenter MB' of the two circles is orthogonal to their radical axis B''A'', which is parallel to the side BA. Thus points O, A' and A of the radius OA, project to points A'', N, A* on BC, A'', A* being symmetric with respect to N. The symmetric H of O with respect to M must, then, lie on the altitude AA*. Since the same is true for all altitudes, this is the orthocenter of ABC. This implies that the parallel MA"' from M to the radius OA defines A''', coinciding with the intersection of the Altitude with the Euler circle and being also the middle of AH.
- The complementary triangle A''B''C'' and A'B'C' have the same Euler circle, whose center Q is the middle of MO. These two triangles are symmetric with respect to Q. Their vertices form a symmetric hexagon A'B''B'A''C'C'', with center at Q. Its area is half the area of the triangle.
- OA'' is equal and parallel to A'M, because of the symmetry of triangles A'B'C' and A''B''C'' with respect to Q. A'M is equal and parallel to HA''', hence OA'' is half the length of HA. Thus, the median AA'' divides HO in ratio 2/1, the division point being the centroid of triangle ABC.
- The orthocenter H of ABC is the [DeLongshamps] point (triangle center X(20)) of triangle A''B''C''.