Vladimir Nikolin

Let BHb and CHc be two altitudes in ΔABC and O its circumcenter. As we know HbHc and AO are perpendicular.

Figure 1

Another proof:

Let H' be the refection of H using line AB. We have, by the reflection property, AH=AH'. But, let H'' be the refection of H using line AC. We have, also by the reflection property, AH=AH'', so AH'=AH'', ΔH'AH'' is isosceles (Blue triangle - figure 1).
As we already know, points H' and H'' (and point A, of course) lies on the circumcircle of the triangle ABC. Its circumcenter is the point O as well and lies on the perpendicular bisector of the line segment H'H'' i.e. H'H'' is perpendicular to AO.
Finally, HbHc is parallel to H'H'' (as midline of the triangle HH'H'') and we conclude that lines HbHc and AO are perpendicular.

Consequence 1:

Let S be the area of the triangle ABC, R the circumradius and po the semiperimeter of the orthic triangle HaHbHc, po = (HaHb + HbHc + HcHa)/2. Then
S = Rpo (only for acute triangles)

Figure 2

Proof:

By Nagel's theorem, the sides of the orthic triangle are perpendicular to the radius-vectors from O to vertices of ΔABC. Which means that the quadrilaterals OHaCHb,OHbAHc and OHcBHa are orthodiagonal quadrilaterals, i.e. all have their diagonals perpendicular. The area of an orthodiagonal quadrilateral, convex or concave, with diagonals d1 and d2 is S = d1d2/2. Now:
Area(ABC) = Area(OHaCHb) + Area(OHbAHc) + Area(OHcBHa)
= RHaHb/2 + RHbHc/2 +RHcHa /2
= R(HaHb + HbHc + HcHa)/2
= Rpo

Remark:

Since all antiparallels to a given side of a triangle are parallel to each other, Nagel's theorem admits am immediate generalization: the line joining the circumcenter to a vertex is perpendicular to the antiparallels to the opposite side. There is a direct proof of that result.

References