Difference between revisions of "Orthocenter"

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''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.''
 
''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.
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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 GO</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.
  
 
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.
 
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 08:30, 5 February 2009

The orthocenter $O$.

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.

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.

See Also