Difference between revisions of "Orthocenter"

 
(Replaced Ceva proof with more elegant Euler line proof)
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The '''orthocenter''' of a [[triangle]] is the point of intersection of its [[altitude]]s.
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The '''orthocenter''' of a [[triangle]] is the point of intersection of its [[altitude]]s. It is conventionally denoted as <math>\displaystyle H</math>.
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== Proof that the altitudes of a triangle are concurrent ==
 
 
[[Image:Orthoproof1.PNG|center]]
 
[[Image:Orthoproof1.PNG|center]]
  
Using the [[trigonometric version of Ceva's Theorem]] it suffices to show that <math>\sin BAD\cdot \sin EBC\cdot  \sin FCE = \sin CAD\cdot \sin ABE\cdot \sin BCF</math>.  Using the right angles gives us
 
  
{| class="wikitable" style="margin: 1em auto 1em auto;height:300px"
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== Proof of Existence ==
| <math>\sin BAD = \frac{HF}{AH}</math>
 
|-
 
| <math>\sin EBC = \frac{HD}{BH}</math>
 
|-
 
| <math>\sin FCE = \frac{HE}{CH}</math>
 
|-
 
| <math>\sin CAD = \frac{HE}{AH}</math>
 
|-
 
| <math>\sin ABE = \frac{HF}{BH}</math>
 
|-
 
| <math>\sin BCF = \frac{HD}{CH}</math>
 
|}
 
  
Thus our previous expression can be rewritten as
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''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 | Euler]], is much more clever, illuminating and insightful.''
  
<center><math> \frac{HF}{AH}\cdot \frac{HD}{BH}\cdot \frac{HE}{CH} = \frac{HE}{AH}\cdot \frac{HF}{BH}\cdot \frac{HD}{CH}. </math></center>
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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 is obviously true so we conclude that the altitudes of a triangle are concurrent.
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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]], after the person to whom this proof is due.

Revision as of 21:03, 25 August 2006

The orthocenter of a triangle is the point of intersection of its altitudes. It is conventionally denoted as $\displaystyle H$.


Orthoproof1.PNG


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 Euler, is much more clever, illuminating and insightful.

Consider a triangle $\displaystyle ABC$ with circumcenter $\displaystyle O$ and centroid $\displaystyle G$. Let $\displaystyle A'$ be the midpoint of $\displaystyle BC$. Let $\displaystyle H$ be the point such that $\displaystyle G$ is between $\displaystyle H$ and $\displaystyle O$ and $\displaystyle HG = 2 HO$. Then the triangles $\displaystyle AGH$, $\displaystyle A'GO$ are similar by angle-side-angle similarity. It follows that $\displaystyle AH$ is parallel to $\displaystyle OA'$ and is therefore perpendicular to $\displaystyle BC$; i.e., it is the altitude from $\displaystyle A$. Similarly, $\displaystyle BH$, $\displaystyle CH$, are the altitudes from $\displaystyle B$, $\displaystyle {C}$. Hence all the altitudes pass through $\displaystyle H$. 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, after the person to whom this proof is due.