Difference between revisions of "Orthocenter"
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== Proof of Existence == | == Proof of Existence == | ||
| − | ''Note: The orthocenter's existence is a trivial consequence of the [[ | + | ''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 <math>ABC</math> with [[circumcenter]] <math>O</math> and [[centroid]] <math>G</math>. Let <math>A'</math> be the midpoint of <math>BC</math>. Let <math>H</math> be the point such that <math>G</math> is between <math>H</math> and <math>O</math> and <math>HG = 2 HO</math>. Then the triangles <math>AGH</math>, <math>A'GO</math> are [[similar]] by angle-side-angle similarity. It follows that <math>AH</math> is parallel to <math>OA'</math> and is therefore perpendicular to <math>BC</math>; i.e., it is the altitude from <math>A</math>. Similarly, <math>BH</math>, <math>CH</math>, are the altitudes from <math>B</math>, <math>{C}</math>. Hence all the altitudes pass through <math>H</math>. Q.E.D. | Consider a triangle <math>ABC</math> with [[circumcenter]] <math>O</math> and [[centroid]] <math>G</math>. Let <math>A'</math> be the midpoint of <math>BC</math>. Let <math>H</math> be the point such that <math>G</math> is between <math>H</math> and <math>O</math> and <math>HG = 2 HO</math>. Then the triangles <math>AGH</math>, <math>A'GO</math> are [[similar]] by angle-side-angle similarity. It follows that <math>AH</math> is parallel to <math>OA'</math> and is therefore perpendicular to <math>BC</math>; i.e., it is the altitude from <math>A</math>. Similarly, <math>BH</math>, <math>CH</math>, are the altitudes from <math>B</math>, <math>{C}</math>. Hence all the altitudes pass through <math>H</math>. Q.E.D. | ||
Revision as of 21:22, 12 December 2007
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 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.
