## Median property

Draw circle k with diameter the side BC of a given triangle ABC. Let the median through A intersect k at point K. Consider the right angled triangles at K: KBB'', KCC", formed by extending CK, BK respectively. Let B* be the middle of BB'' and draw from B a parallel to B*K intersecting the median at L. Show that L coincides with the projection of the orthocenter H on the median.

Corollary The three points La, Lb, Lc constructed on the three medians respectively with the above procedure lie on the orthocentroidal circle of the triangle (the circle with diameter HG, H: orthocenter, G: barycenter, see second figure below).

[1] Triangles BCL and B*KC* are homothetic. Measuring the angle at K gives:
chi + omega = A + pi/2.
angle(B*KC*) = angle( BLC) = pi/2 + pi - (chi+omega) = pi - A.
Since also angle(BHC) = pi - A, {B,H,L,C} are on a circle c'.
[2] By well known property of the orthocenter H, c' coincides with the reflected on BC of the circumcircle of triangle ABC.
[3] This implies that the intersection point L' of AM with c' is the symmetric of A with respect to M and the intersection point of AH with c' is the reflected point of A with respect to BC. Thus angle(HH'L') is a right one and since HKL'H' is cyclic angle(HKL'), opposite to H', is also a right one.

Triangle LaLbLc, with point Li constructed on each median with the procedure described above, is similar to the triangle of medians DBMb (the triangle whose sides are the medians of ABC).