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.

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