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. | ||
| + | |||
| + | == 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>{A'}</math>. Construct <math>\displaystyle{B'}</math> in a similar way extending <math>BY</math>. | ||
| + | <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: | ||
| + | <math>\begin{displaymath}\frac{BX}{XC}=\frac{AB}{CA'},\qquad\frac{CY}{YA}=\frac{CB'}{BA}\end{displaymath}</math> | ||
| + | |||
| + | 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> | ||
| + | |||
| + | 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>. | ||
| + | |||
| + | 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> | ||
| + | |||
| + | which lead to | ||
| + | <math>\begin{displaymath}\frac{AZ}{ZB}=\frac{A'C}{CB'}.\end{displaymath}</math> | ||
| + | |||
| + | 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> | ||
| + | |||
| + | and we conclude the proof. | ||
| + | |||
| + | 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> | ||
| + | |||
| + | 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> | ||
| + | |||
| + | and thus | ||
| + | <math>\begin{displaymath}\frac{AZ'}{Z'B}=\frac{AZ}{ZB}\end{displaymath}</math> | ||
| + | |||
| + | 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.
Contents
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
where all segments in the formula are directed segments.
Proof
Let 
be points on 
respectively such that 
are concurrent, and let 
be the point where 
, 
and 
meet. Draw a parallel to 
through the point 
. Extend until it intersects the parallel at a point 
. Construct 
in a similar way extending .
The triangles 
are similar, and so are 
and 
. 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. Unknown error_msg)
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. Unknown error_msg)
Notice that if directed segments are being used then and 
have opposite signs, and therefore when cancelled change the sign of the expression. That's why we changed 
to 
.
Now we turn to consider the following similarities: 
and 
. 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. Unknown error_msg)
which lead to $\begin{displaymath}\frac{AZ}{ZB}=\frac{A'C}{CB'}.\end{displaymath}$ (Error compiling LaTeX. Unknown error_msg)
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. Unknown error_msg)
and we conclude the proof.
To prove the converse, suppose that 
are points on 
respectively and satisfying
$\begin{displaymath}\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1.\end{displaymath}$ (Error compiling LaTeX. Unknown error_msg)
Let 
be the intersection point of with , and let 
be the intersection of 
with . Since then 
are concurrent, we have
$\begin{displaymath}\frac{AZ'}{Z'B}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1\end{displaymath}$ (Error compiling LaTeX. Unknown error_msg)
and thus $\begin{displaymath}\frac{AZ'}{Z'B}=\frac{AZ}{ZB}\end{displaymath}$ (Error compiling LaTeX. Unknown error_msg)
which implies 
, and therefore are concurrent.
Example
Suppose AB, AC, and BC have lengths 13, 14, and 15. If 
and 
. Find BD and DC.
If 
and 
, then 
, and 
. From this, we find 
and 
.
