# Difference between revisions of "Orthocenter"

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Let the line through <math>B</math> parallel to <math>AC</math> and the line through <math>C</math> parallel to <math>AB</math> intersect at <math>D.</math> Define <math>E,F</math> similarly. Note that <math>FA=BC=AE,</math> so the <math>A</math> altitude of <math>\triangle ABC</math> is the perpendicular bisector of <math>DE.</math> Since the circumcenter exists, the altitude must too. | Let the line through <math>B</math> parallel to <math>AC</math> and the line through <math>C</math> parallel to <math>AB</math> intersect at <math>D.</math> Define <math>E,F</math> similarly. Note that <math>FA=BC=AE,</math> so the <math>A</math> altitude of <math>\triangle ABC</math> is the perpendicular bisector of <math>DE.</math> Since the circumcenter exists, the altitude must too. | ||

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import olympiad; | import olympiad; | ||

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label("F",F,dir(140)); | label("F",F,dir(140)); | ||

label("H",H,dir(50)); | label("H",H,dir(50)); | ||

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==Properties== | ==Properties== |

## Latest revision as of 20:52, 3 May 2020

The **orthocenter** of a triangle is the point of intersection of its altitudes. It is conventionally denoted .
The lines highlighted are the altitudes of the triangle, they meet at the orthocenter.

## 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.*

Consider a triangle with circumcenter and centroid . Let be the midpoint of . Let be the point such that is between and and . Then the triangles , are similar by side-angle-side similarity. It follows that is parallel to and is therefore perpendicular to ; i.e., it is the altitude from . Similarly, , , are the altitudes from , . Hence all the altitudes pass through . 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, and also contains the triangle's de Longchamps point and nine-point center.

### Easier proof

That seems somewhat overkill to prove the existence of the orthocenter. We use a much easier (and funnier) way.

Let the line through parallel to and the line through parallel to intersect at Define similarly. Note that so the altitude of is the perpendicular bisector of Since the circumcenter exists, the altitude must too.

## Properties

- The orthocenter and the circumcenter of a triangle are isogonal conjugates.
- If the orthocenter's triangle is acute, then the orthocenter is in the triangle; if 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 be a triangle and its orthocenter. Then the reflections of over , , and are on the circumcircle of :

- Even more interesting is the fact that if you take any point on the circumcircle and let to be the midpoint of , then is on the nine-point circle.