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Orthocenter

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The orthocenter of a triangle is the point of intersection of its altitudes. It is conventionally denoted H.

Proof of Existence

Note: The orthocenter's existence is a trivial consequence of the trigonometric version Ceva's Theorem; however, the following proof, due to Leonhard Euler, is much more clever, illuminating and insightful.

defaultpen(fontsize(8));pair A=(8,7), B=(0,0), C=(10,0), A1 = (B+C)/2, O = circumcenter(A,B,C), G = (A+B+C)/3, H = 3*G-2*O;dr... Consider a triangle ABC with circumcenter O and centroid G. Let A' be the midpoint of BC. Let H be the point such that G is between H and O and HG = 2 GO. Then the triangles AGH, A'GO are similar by angle-side-angle similarity. It follows that AH is parallel to OA' and is therefore perpendicular to BC; i.e., it is the altitude from A. Similarly, BH, CH, are the altitudes from B, {C}. Hence all the altitudes pass through H. Q.E.D.

This proof also gives us the result that the orthocenter, centroid, and circumcenter are collinear, in that order, and in the proportions described above. The line containing these three points is known as the Euler line of the triangle.

Properties

  • The orthocenter is collinear with the circumcenter and de Longchamps point.
  • If the orthocenter's triangle is acute, then the orthocenter is on the triangle, it the triangle is right, then it is on the vertex opposite the hypotenuse, and if it is obtuse, then the orthocenter is outside the triangle.
  • Let ABC be a triangle and H its orthocenter. Then the reflections of H over AB, BC, and CA are on the circumcircle of ABC:

defaultpen(fontsize(8));pair A=(8,7), B=(0,0), C=(10,0), H=orthocenter(A,B,C), A1, B1, C1;A1 = 2*foot(A,B,C)-H;B1 = 2*foot(B,...

See Also

Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!
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