Vladimir Nikolin

    Nagel's Theorem claims that line HbHc (feets of the heights hb and hc) is peprendicular to the radius-vector of the circumcircle drawn to the corresponding vertex A.

    More generally: Let D be the point on the side BC and E be the point on the side AC so that ∠AEB = ∠ADE = φ. Line DE is perpendicular to the radius-vector CO of the circumcircle.

 
 


Figure 1

    Proof: Let P be the point of the intersection of the lines DE and CO and F be the point on the circumcircle such that CF is a diameter of the circumcircle of ΔABC.

  • First, notice that the quadrilateral ABDE is cyclic (see here), so ∠ABD + ∠AED = 180.
  • Second, ∠AFB = ∠ABC, as inscribed angles subtend by the same arc AC.
  • Third, since CF is a diameter, ∠CAF = 90 (as an inscribed angle subtended by a diameter).
  • Finally, the quadrilateral AFPE is also cyclic (see first and second, together), so ∠EPF = 180 - ∠CAF = 90.

 
    Nagel's Theorem is just a special case, where ∠AEB = ∠ADE = 90. Also, allowed is that points E and D can be outside of the sides BC and AC, but proof is similar and little complicated. An interesting applet ilustrated this.

    Consequence: Let Γ be the circle passing through A and B. Suppose Γ intersects the line AC at X and intersects the line BC at Y. Line XY is now perpendicular to OC, where O is circumcenter of the triangle ABC.