MedianProperty

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 L_a, L_b, L_c 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 L_aL_bL_c, with point L_i constructed on each median with the procedure described above, is similar to the triangle of medians DBM_b (the triangle whose sides are the medians of ABC).

Median property

See Also