Let the three altitudes of ⊿ABC be AD, BE, and CF, among which D, E, and F are the feet of perpendiculars, and the orthocenter is H. The opposite sides of angles A, B, and C are a, b, and c respectively.

1. The orthocenter of an acute triangle is inside the triangle; the orthocenter of a right triangle is at the vertex of the right angle; the orthocenter of an obtuse triangle is outside the triangle.

2. The orthocenter of a triangle is the incenter of its foot triangle; or, the incenter of a triangle is the orthocenter of its paracentric triangle; 3. The symmetric points of the orthocenter H about the three sides are all on the circumcircle of △ABC.

4. In △ABC, there are six groups of four points that are cocircular, there are three groups (four in each group) of similar right triangles, and AH·HD=BH·HE=CH·HF.

5. Any point among H, A, B, and C is the orthocenter of a triangle with the other three points as vertices (such four points are called an orthocenter group).

6. The circumscribed circles of △ABC, △ABH, △BCH, and △ACH are equal circles.

7. In a non-right triangle, the straight line passing through H intersects the straight lines AB and AC at P and Q respectively, then AB/AP·tanB+AC/AQ·tanC=tanA+tanB+tanC.

8. The distance from any vertex of a triangle to the orthocenter is equal to twice the distance from the circumcenter to the opposite side.

9. Let O and H be the circumcenter and orthocenter of △ABC respectively, then ∠BAO=∠HAC, ∠ABH=∠OBC, ∠BCO=∠HCA.

10. The sum of the distances from the orthocenter of an acute triangle to its three vertices is equal to twice the sum of the radii of its incircle and circumcircle.

11. The orthocenter of an acute triangle is the incenter of the foot triangle; among the inscribed triangles of an acute triangle (vertices are on the sides of the original triangle), the perimeter of the foot triangle is the shortest.

12. Simson's Theorem (Simson Line): The necessary and sufficient condition for the feet of perpendicular lines drawn from a point to the three sides of a triangle to be collinear is that the point falls on the circumcircle of the triangle.

13. Suppose there is a point P inside the acute angle ⊿ABC, then the necessary and sufficient condition for P to be the orthocenter is PB*PC*BC+PB*PA*AB+PA*PC*AC=AB*BC*CA.