*Vladimir Nikolin*

**Nagel's Theorem** claims that line H_{b}H_{c} (feets of the heights h_{b} and h_{c}) 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.

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.