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 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.''
  
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==Properties==
 
==Properties==
 
*The orthocenter is collinear with the [[circumcenter]] and [[de Longchamps point]].
 
*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 triangle|right]], then it is on the vertex opposite the [[hypotenuse]], and if it is [[obtuse]], then the orthocenter is outside the triangle.
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*If the orthocenter's triangle is [[acute triangle|acute]], then the orthocenter is on the triangle, it the triangle is [[right triangle|right]], then it is on the vertex opposite the [[hypotenuse]], and if it is [[obtuse triangle|obtuse]], then the orthocenter is outside the triangle.
  
 
==See Also==
 
==See Also==

Revision as of 21:49, 20 April 2008

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 $HG = 2 HO$. 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