Difference between revisions of "Ceva's Theorem"

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<br><center><math>BD * CE * AF = +DC * EA * FB</math></center><br>
 
<br><center><math>BD * CE * AF = +DC * EA * FB</math></center><br>
 
where all segments in the formula are directed segments.
 
where all segments in the formula are directed segments.
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== 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>{A'}</math>. Construct <math>\displaystyle{B'}</math> in a similar way extending <math>BY</math>.
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<center>''(ceva1.png)''</center>
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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:
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<math>\begin{displaymath}\frac{BX}{XC}=\frac{AB}{CA'},\qquad\frac{CY}{YA}=\frac{CB'}{BA}\end{displaymath}</math>
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and thus
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<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>
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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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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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<math>\begin{displaymath}\frac{CP}{ZP}=\frac{A'C}{AZ},\qquad\frac{CP}{ZP}=\frac{CB'}{ZB}\end{displaymath}</math>
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which lead to
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<math>\begin{displaymath}\frac{AZ}{ZB}=\frac{A'C}{CB'}.\end{displaymath}</math>
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Multiplying the last expression with (1) gives
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<math>\begin{displaymath}\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1\end{displaymath}</math>
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and we conclude the proof.
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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>
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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
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<math>\begin{displaymath}\frac{AZ'}{Z'B}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1\end{displaymath}</math>
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 +
and thus
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<math>\begin{displaymath}\frac{AZ'}{Z'B}=\frac{AZ}{ZB}\end{displaymath}</math>
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which implies <math>Z=Z'</math>, and therefore <math>AX,BY,CZ</math> are concurrent.
  
 
== Example ==
 
== Example ==

Revision as of 16:36, 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 ${A'}$. Construct $\displaystyle{B'}$ in a similar way extending $BY$.

(ceva1.png)

The triangles $\triangle{ABX}<math> and <math>\triangle{A'CX}$ are similar, and so are $\triangle{ABY}$ and $\triangle{CB'Y}$. Then the following equalities hold: $\begin{displaymath}\frac{BX}{XC}=\frac{AB}{CA'},\qquad\frac{CY}{YA}=\frac{CB'}{BA}\end{displaymath}$ (Error compiling LaTeX. ! LaTeX Error: Bad math environment delimiter.)

and thus $\begin{displaymath} \frac{BX}{XC}\cdot\frac{CY}{YA}=\frac{AB}{CA'}\cdot\frac{CB'}{BA}=\frac{CB'}{A'C}. \end{displaymath} (1)$ (Error compiling LaTeX. ! LaTeX Error: Bad math environment delimiter.)

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 $\begin{displaymath}\frac{CP}{ZP}=\frac{A'C}{AZ},\qquad\frac{CP}{ZP}=\frac{CB'}{ZB}\end{displaymath}$ (Error compiling LaTeX. ! LaTeX Error: Bad math environment delimiter.)

which lead to $\begin{displaymath}\frac{AZ}{ZB}=\frac{A'C}{CB'}.\end{displaymath}$ (Error compiling LaTeX. ! LaTeX Error: Bad math environment delimiter.)

Multiplying the last expression with (1) gives $\begin{displaymath}\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1\end{displaymath}$ (Error compiling LaTeX. ! LaTeX Error: Bad math environment delimiter.)

and we conclude the proof.

To prove the converse, suppose that $X,Y,Z$ are points on ${BC, CA, AB}$ respectively and satisfying $\begin{displaymath}\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1.\end{displaymath}$ (Error compiling LaTeX. ! LaTeX Error: Bad math environment delimiter.)

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 $\begin{displaymath}\frac{AZ'}{Z'B}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1\end{displaymath}$ (Error compiling LaTeX. ! LaTeX Error: Bad math environment delimiter.)

and thus $\begin{displaymath}\frac{AZ'}{Z'B}=\frac{AZ}{ZB}\end{displaymath}$ (Error compiling LaTeX. ! LaTeX Error: Bad math environment delimiter.)

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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