Difference between revisions of "Ceva's Theorem"

(Statement)
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== Statement ==
 
== Statement ==
http://billydorminy.homelinux.com/aopswiki/cevathm.png
 
 
 
A [[necessary and sufficient]] condition for <math>AD, BE, CF,</math> where <math>D, E,</math> and <math>F</math> are points of the respective side lines <math>BC, CA, AB</math> of a triangle <math>ABC</math>, to be concurrent is that
 
A [[necessary and sufficient]] condition for <math>AD, BE, CF,</math> where <math>D, E,</math> and <math>F</math> are points of the respective side lines <math>BC, CA, AB</math> of a triangle <math>ABC</math>, to be concurrent is that
 
<br><center><math>BD\cdot CE\cdot AF = DC \cdot EA \cdot FB</math></center><br>
 
<br><center><math>BD\cdot CE\cdot AF = DC \cdot EA \cdot FB</math></center><br>
 
where all segments in the formula are [[directed segments]].
 
where all segments in the formula are [[directed segments]].
 +
 +
[[Image:Ceva1.PNG|center]]
  
 
== Proof ==
 
== Proof ==
Let <math>{X,Y,Z}</math> be points on <math>{BC}, {CA}, {AB}</math> respectively such that <math>AX,BY,CZ</math> are concurrent, and let <math>{P}</math> be the point where <math>AX</math>, <math>BY</math> and <math>CZ</math> meet. Draw a parallel to <math>AB</math> through the point <math>{C}</math>. Extend <math>AX</math> until it intersects the parallel at a point <math>\displaystyle{A'}</math>. Construct <math>\displaystyle{B'}</math> in a similar way extending <math>BY</math>.
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Letting the [[altitude]] from <math>A</math> to <math>BC</math> have length <math>h</math> we have <math>[ABD]=\frac 12 BD\cdot h</math> and <math>[ACD]=\frac 12 DC\cdot h</math> where the brackets represent [[area]]. Thus <math>\frac{[ABD]}{[ACD]} = \frac{BD}{DC}</math>. In the same manner, we find that <math>\frac{[XBD]}{[XCD]} = \frac{BD}{DC}</math>.  Thus <center><math> \frac{BD}{DC} = \frac{[ABD]}{[ACD]} = \frac{[XBD]}{[XCD]} = \frac{[ABD]-[XBD]}{[ACD]-[XCD]} = \frac{[ABX]}{[ACX]}. </math></center>
<center>''(ceva1.png)''</center>
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The triangles <math>\displaystyle{\triangle{ABX}}</math> and <math>\displaystyle{\triangle{A'CX}}</math> are similar, and so are <math>\displaystyle\triangle{ABY}</math> and <math>\triangle{CB'Y}</math>. Then the following equalities hold:
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Likewise, we find that
<center><math>\frac{BX}{XC}=\frac{AB}{CA'},\qquad\frac{CY}{YA}=\frac{CB'}{BA}</math></center>
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<br>
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{| class="wikitable" style="margin: 1em auto 1em auto;height:100px"
and thus
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| <math>\frac{CE}{EA}</math> || <math>=\frac{[BCX]}{[ABX]}</math>
<center><math>\frac{BX}{XC}\cdot\frac{CY}{YA}=\frac{AB}{CA'}\cdot\frac{CB'}{BA}=\frac{CB'}{A'C} \qquad(1)</math></center>
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|-
<br>
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| <math>\frac{AF}{FB}</math> || <math>=\frac{[ACX]}{[BCX]}</math>
Notice that if directed segments are being used, then <math>AB</math> and <math>BA</math> have opposite signs, and therefore when cancelled change the sign of the expression. That's why we changed <math>CA'</math> to <math>A'C</math>.
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|}
<br><br>
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Now we turn to consider the following similarities: <math>\triangle{AZP}\sim\triangle{A'CP}</math> and <math>\triangle BZP\sim\triangle B'CP</math>. From them we get the equalities
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Thus <center><math> \frac{BD}{DC}\cdot \frac{CE}{EA}\cdot \frac{AF}{FB} = \frac{[ABX]}{[ACX]}\cdot \frac{[BCX]}{[ABX]}\cdot \frac{[ACX]}{[BCX]} = 1 \Rightarrow BD\cdot CE\cdot AF = DC \cdot EA \cdot FB. </math></center>
<center><math>\frac{CP}{ZP}=\frac{A'C}{AZ},\qquad\frac{CP}{ZP}=\frac{CB'}{ZB}</math></center>
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<br>
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<math>\mathcal{QED}</math>
which lead to
 
<center><math>\frac{AZ}{ZB}=\frac{A'C}{CB'}</math>.</center>
 
<br>
 
Multiplying the last expression with (1) gives
 
<center><math>\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1</math></center>
 
<br>
 
and we conclude the proof.
 
<br><br>
 
To prove the converse, suppose that <math>{X,Y,Z}</math> are points on <math>{BC}, {CA}, {AB}</math> respectively and satisfying
 
<center><math>\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1.</math></center>
 
<br>
 
Let <math>Q</math> be the intersection point of <math>AX</math> with <math>BY</math>, and let <math>Z'</math> be the intersection of <math>CQ</math> with <math>AB</math>. Since then <math>AX,BY,CZ'</math> are concurrent, we have
 
<center><math>\frac{AZ'}{Z'B}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1</math></center>
 
<br>
 
and thus
 
<center><math>\frac{AZ'}{Z'B}=\frac{AZ}{ZB}</math></center>
 
<br>
 
which implies <math>Z=Z'</math>, and therefore <math>AX,BY,CZ</math> are concurrent.
 
  
<div align="right">''(proof courtesy planetmath.org, used under GNU License)''</div>
 
  
== Example ==
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== Examples ==
Suppose AB, AC, and BC have lengths 13, 14, and 15.  If <math>\frac{AF}{FB} = \frac{2}{5}</math> and <math>\frac{CE}{EA} = \frac{5}{8}</math>.  Find BD and DC.<br>
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# Suppose AB, AC, and BC have lengths 13, 14, and 15.  If <math>\frac{AF}{FB} = \frac{2}{5}</math> and <math>\frac{CE}{EA} = \frac{5}{8}</math>.  Find BD and DC.<br> <br> If <math>BD = x</math> and <math>DC = y</math>, then <math>10x = 40y</math>, and <math>{x + y = 15}</math>.  From this, we find <math>x = 12</math> and <math>y = 3</math>.
<br>
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# See the proof of the concurrency of the altitudes of a triangle at the [[orthocenter]].
If <math>BD = x</math> and <math>DC = y</math>, then <math>10x = 40y</math>, and <math>{x + y = 15}</math>.  From this, we find <math>x = 12</math> and <math>y = 3</math>.
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# See the proof of the concurrency of the perpendicual bisectors of a triangle at the [[circumcenter]].
  
 
== See also ==
 
== See also ==
 
* [[Menelaus' Theorem]]
 
* [[Menelaus' Theorem]]
 
* [[Stewart's Theorem]]
 
* [[Stewart's Theorem]]

Revision as of 11:03, 18 August 2006

Ceva's Theorem is an algebraic statement regarding the lengths of cevians in a triangle.


Statement

A necessary and sufficient condition for $AD, BE, CF,$ where $D, E,$ and $F$ are points of the respective side lines $BC, CA, AB$ of a triangle $ABC$, to be concurrent is that


$BD\cdot CE\cdot AF = DC \cdot EA \cdot FB$


where all segments in the formula are directed segments.

Ceva1.PNG

Proof

Letting the altitude from $A$ to $BC$ have length $h$ we have $[ABD]=\frac 12 BD\cdot h$ and $[ACD]=\frac 12 DC\cdot h$ where the brackets represent area. Thus $\frac{[ABD]}{[ACD]} = \frac{BD}{DC}$. In the same manner, we find that $\frac{[XBD]}{[XCD]} = \frac{BD}{DC}$. Thus

$\frac{BD}{DC} = \frac{[ABD]}{[ACD]} = \frac{[XBD]}{[XCD]} = \frac{[ABD]-[XBD]}{[ACD]-[XCD]} = \frac{[ABX]}{[ACX]}.$

Likewise, we find that

$\frac{CE}{EA}$ $=\frac{[BCX]}{[ABX]}$
$\frac{AF}{FB}$ $=\frac{[ACX]}{[BCX]}$

Thus

$\frac{BD}{DC}\cdot \frac{CE}{EA}\cdot \frac{AF}{FB} = \frac{[ABX]}{[ACX]}\cdot \frac{[BCX]}{[ABX]}\cdot \frac{[ACX]}{[BCX]} = 1  \Rightarrow BD\cdot CE\cdot AF = DC \cdot EA \cdot FB.$

$\mathcal{QED}$


Examples

  1. Suppose AB, AC, and BC have lengths 13, 14, and 15. If $\frac{AF}{FB} = \frac{2}{5}$ and $\frac{CE}{EA} = \frac{5}{8}$. Find BD and DC.

    If $BD = x$ and $DC = y$, then $10x = 40y$, and ${x + y = 15}$. From this, we find $x = 12$ and $y = 3$.
  2. See the proof of the concurrency of the altitudes of a triangle at the orthocenter.
  3. See the proof of the concurrency of the perpendicual bisectors of a triangle at the circumcenter.

See also

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