[alogo] On the medial lines of triangle-sides

Consider the medial line DH, of side BC, of triangle ABC. Let K, G be its intersection-points with the sides AB, AC respectively. The median AH of triangle AGK, intersects the circumcircle at a point I, such that DI is orthogonal to IH and contains the orthocenter of the triangle ABC.

[0_0] [0_1]
[1_0] [1_1]
[2_0] [2_1]

Consider the circumcircle (c) of the rectangle BDGF. Let I be the intersection point of HF with this circle. Since FD is a diameter of (c), the angle at I is a right one. Let L be the circumcenter of ABC and let N be the intersection-point of AL and ID. Then angle(BIN) = angle(BID) = angle(BGD) = (pi/2-angle(C)) = angle(BAN). This implies that points B, I, A and N are on the same circle and that AN is a diameter of the circumcircle (d) of ABC.
For the location of the orthocenter, consider M, the intersection point of NI and the altitude AE. Then since L is the middle of AN, LD is half the length of AM. But this is identifies M as the orthocenter.
Notice, that, by its construction, FH is harmonic conjugate to the altitude AE, with respect to the sides AB, AC.
Line IN and the point I are relate to the parabola tangent to the three sides of the triangle and the medial line GD. Look at MedialParabola.html for the details.

Produced with EucliDraw©