Difference between revisions of "Ceva's Theorem"
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== 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>{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>{\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> | <math>\begin{displaymath}\frac{BX}{XC}=\frac{AB}{CA'},\qquad\frac{CY}{YA}=\frac{CB'}{BA}\end{displaymath}</math> | ||
Revision as of 16:38, 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
![$BD * CE * AF = +DC * EA * FB$](http://latex.artofproblemsolving.com/d/0/e/d0e1f991c44cbb04a7373db29ab9c1bdabff6d2b.png)
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 and
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
.