Difference between revisions of "Orthocenter"
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| − | The '''orthocenter''' of a [[triangle]] is the point of intersection of its [[altitude]]s. It is conventionally denoted | + | The '''orthocenter''' of a [[triangle]] is the point of intersection of its [[altitude]]s. It is [[mathematical convention | conventionally]] denoted <math>\displaystyle H</math>. |
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== Proof of Existence == | == Proof of Existence == | ||
| − | ''Note: The orthocenter's existence is a trivial consequence of the [[trigonometric version of Ceva's Theorem]]; however, the following proof, due to [[Leonhard | + | ''Note: The orthocenter's existence is a trivial consequence of the [[trigonometric version of Ceva's Theorem]]; however, the following proof, due to [[Leonhard Euler]], is much more clever, illuminating and insightful.'' |
Consider a triangle <math>\displaystyle ABC</math> with [[circumcenter]] <math>\displaystyle O</math> and [[centroid]] <math>\displaystyle G</math>. Let <math>\displaystyle A'</math> be the midpoint of <math>\displaystyle BC</math>. Let <math>\displaystyle H</math> be the point such that <math>\displaystyle G</math> is between <math>\displaystyle H</math> and <math>\displaystyle O</math> and <math>\displaystyle HG = 2 HO</math>. Then the triangles <math>\displaystyle AGH</math>, <math>\displaystyle A'GO</math> are [[similar]] by angle-side-angle similarity. It follows that <math>\displaystyle AH</math> is parallel to <math>\displaystyle OA'</math> and is therefore perpendicular to <math>\displaystyle BC</math>; i.e., it is the altitude from <math>\displaystyle A</math>. Similarly, <math>\displaystyle BH</math>, <math>\displaystyle CH</math>, are the altitudes from <math>\displaystyle B</math>, <math>\displaystyle {C}</math>. Hence all the altitudes pass through <math>\displaystyle H</math>. Q.E.D. | Consider a triangle <math>\displaystyle ABC</math> with [[circumcenter]] <math>\displaystyle O</math> and [[centroid]] <math>\displaystyle G</math>. Let <math>\displaystyle A'</math> be the midpoint of <math>\displaystyle BC</math>. Let <math>\displaystyle H</math> be the point such that <math>\displaystyle G</math> is between <math>\displaystyle H</math> and <math>\displaystyle O</math> and <math>\displaystyle HG = 2 HO</math>. Then the triangles <math>\displaystyle AGH</math>, <math>\displaystyle A'GO</math> are [[similar]] by angle-side-angle similarity. It follows that <math>\displaystyle AH</math> is parallel to <math>\displaystyle OA'</math> and is therefore perpendicular to <math>\displaystyle BC</math>; i.e., it is the altitude from <math>\displaystyle A</math>. Similarly, <math>\displaystyle BH</math>, <math>\displaystyle CH</math>, are the altitudes from <math>\displaystyle B</math>, <math>\displaystyle {C}</math>. Hence all the altitudes pass through <math>\displaystyle H</math>. 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]] | + | 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. |
Revision as of 10:28, 7 September 2006
The orthocenter of a triangle is the point of intersection of its altitudes. It is conventionally denoted 
.
Proof of Existence
Note: The orthocenter's existence is a trivial consequence of the trigonometric version of 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 angle-side-angle 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.
