Difference between revisions of "Ceva's Theorem"
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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> | + | <center><math>\frac{BX}{XC}=\frac{AB}{CA'},\qquad\frac{CY}{YA}=\frac{CB'}{BA}</math></center> |
− | + | <br> | |
and thus | and thus | ||
− | <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> | + | <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> | + | <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> | + | <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> | + | <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> | + | <center><math>\frac{AZ'}{Z'B}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1</math></center> |
− | + | <br> | |
and thus | and thus | ||
− | <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.
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:
![$\frac{BX}{XC}=\frac{AB}{CA'},\qquad\frac{CY}{YA}=\frac{CB'}{BA}$](http://latex.artofproblemsolving.com/b/0/0/b0082a0befdf70df29ff5444e3acf26efd2e560d.png)
and thus
![$\frac{BX}{XC}\cdot\frac{CY}{YA}=\frac{AB}{CA'}\cdot\frac{CB'}{BA}=\frac{CB'}{A'C} \qquad(1)$](http://latex.artofproblemsolving.com/7/9/a/79a4b058d864397686f7a0aa1e17bec75d9c297e.png)
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
![$\frac{CP}{ZP}=\frac{A'C}{AZ},\qquad\frac{CP}{ZP}=\frac{CB'}{ZB}$](http://latex.artofproblemsolving.com/6/d/a/6dafa6105a85e0b6ddb894c1414b4f58583cbb9b.png)
which lead to
![$\frac{AZ}{ZB}=\frac{A'C}{CB'}$](http://latex.artofproblemsolving.com/3/a/e/3ae1a2b95a62b8e69259d021717176eed6a9297b.png)
Multiplying the last expression with (1) gives
![$\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1$](http://latex.artofproblemsolving.com/4/9/3/493a1d0a84e5a535246f53f4f9a9292d8617f2a3.png)
and we conclude the proof.
To prove the converse, suppose that are points on
respectively and satisfying
![$\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1.$](http://latex.artofproblemsolving.com/8/d/9/8d96c79d368e98538b8db155a11c6b0c4568f14e.png)
Let be the intersection point of
with
, and let
be the intersection of
with
. Since then
are concurrent, we have
![$\frac{AZ'}{Z'B}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1$](http://latex.artofproblemsolving.com/e/2/5/e253fa561bc7d3a9ad4c56cc24a4fd80ac3fb2cc.png)
and thus
![$\frac{AZ'}{Z'B}=\frac{AZ}{ZB}$](http://latex.artofproblemsolving.com/0/d/e/0de813256e2d4f6c06aaf49c40aad01cb9b39d10.png)
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
.