Difference between revisions of "Ceva's Theorem"

(Proof)
(Proof)
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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>.
 
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>.
 
<center>''(ceva1.png)''</center>
 
<center>''(ceva1.png)''</center>
The triangles <math>{\triangle{ABX}}</math> and <math>{\triangle{A'CX}}</math> are similar, and so are <math>\triangle{ABY}</math> and <math>\triangle{CB'Y}</math>. Then the following equalities hold:
+
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:
<math>\begin{displaymath}\frac{BX}{XC}=\frac{AB}{CA'},\qquad\frac{CY}{YA}=\frac{CB'}{BA}\end{displaymath}</math>
+
<center><math>\frac{BX}{XC}=\frac{AB}{CA'},\qquad\frac{CY}{YA}=\frac{CB'}{BA}</math></center>
 
+
<br>
 
and thus
 
and thus
<math>\begin{displaymath} \frac{BX}{XC}\cdot\frac{CY}{YA}=\frac{AB}{CA'}\cdot\frac{CB'}{BA}=\frac{CB'}{A'C}. \end{displaymath} (1)</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>
 
+
<br>
 
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>.
 
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>.
 
+
<br><br>
 
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
 
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
<math>\begin{displaymath}\frac{CP}{ZP}=\frac{A'C}{AZ},\qquad\frac{CP}{ZP}=\frac{CB'}{ZB}\end{displaymath}</math>
+
<center><math>\frac{CP}{ZP}=\frac{A'C}{AZ},\qquad\frac{CP}{ZP}=\frac{CB'}{ZB}</math></center>
 
+
<br>
 
which lead to
 
which lead to
<math>\begin{displaymath}\frac{AZ}{ZB}=\frac{A'C}{CB'}.\end{displaymath}</math>
+
<center><math>\frac{AZ}{ZB}=\frac{A'C}{CB'}</math>.</center>
 
+
<br>
 
Multiplying the last expression with (1) gives
 
Multiplying the last expression with (1) gives
<math>\begin{displaymath}\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1\end{displaymath}</math>
+
<center><math>\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1</math></center>
 
+
<br>
 
and we conclude the proof.
 
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
+
To prove the converse, suppose that <math>{X,Y,Z}</math> are points on <math>{BC}, {CA}, {AB}</math> respectively and satisfying
<math>\begin{displaymath}\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1.\end{displaymath}</math>
+
<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
 
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
<math>\begin{displaymath}\frac{AZ'}{Z'B}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1\end{displaymath}</math>
+
<center><math>\frac{AZ'}{Z'B}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1</math></center>
 
+
<br>
 
and thus
 
and thus
<math>\begin{displaymath}\frac{AZ'}{Z'B}=\frac{AZ}{ZB}\end{displaymath}</math>
+
<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.
 
which implies <math>Z=Z'</math>, and therefore <math>AX,BY,CZ</math> are concurrent.
  

Revision as of 17:01, 20 June 2006

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


Statement

(awaiting image)
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


$BD * CE * AF = +DC * EA * FB$


where all segments in the formula are directed segments.

Proof

Let ${X,Y,Z}$ be points on ${BC}, {CA}, {AB}$ respectively such that $AX,BY,CZ$ are concurrent, and let ${P}$ be the point where $AX$, $BY$ and $CZ$ meet. Draw a parallel to $AB$ through the point ${C}$. Extend $AX$ until it intersects the parallel at a point $\displaystyle{A'}$. Construct $\displaystyle{B'}$ in a similar way extending $BY$.

(ceva1.png)

The triangles $\displaystyle{\triangle{ABX}}$ and $\displaystyle{\triangle{A'CX}}$ are similar, and so are $\displaystyle\triangle{ABY}$ and $\triangle{CB'Y}$. Then the following equalities hold:

$\frac{BX}{XC}=\frac{AB}{CA'},\qquad\frac{CY}{YA}=\frac{CB'}{BA}$


and thus

$\frac{BX}{XC}\cdot\frac{CY}{YA}=\frac{AB}{CA'}\cdot\frac{CB'}{BA}=\frac{CB'}{A'C} 	\qquad(1)$


Notice that if directed segments are being used then $AB$ and $BA$ have opposite signs, and therefore when cancelled change the sign of the expression. That's why we changed $CA'$ to $A'C$.

Now we turn to consider the following similarities: $\triangle{AZP}\sim\triangle{A'CP}$ and $\triangle BZP\sim\triangle B'CP$. From them we get the equalities

$\frac{CP}{ZP}=\frac{A'C}{AZ},\qquad\frac{CP}{ZP}=\frac{CB'}{ZB}$


which lead to

$\frac{AZ}{ZB}=\frac{A'C}{CB'}$.


Multiplying the last expression with (1) gives

$\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1$


and we conclude the proof.

To prove the converse, suppose that ${X,Y,Z}$ are points on ${BC}, {CA}, {AB}$ respectively and satisfying

$\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1.$


Let $Q$ be the intersection point of $AX$ with $BY$, and let $Z'$ be the intersection of $CQ$ with $AB$. Since then $AX,BY,CZ'$ are concurrent, we have

$\frac{AZ'}{Z'B}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1$


and thus

$\frac{AZ'}{Z'B}=\frac{AZ}{ZB}$


which implies $Z=Z'$, and therefore $AX,BY,CZ$ are concurrent.

Example

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 $y = 3$.

See also

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