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Difference between revisions of "Ceva's Theorem"

(Missing dollar sign; didn't actually revise content)
(restated theorem in conventional (and, in my opinion, superior) way; rewrote first proof; also wrote proof for trig version (I think Law of Sines is a dead end))
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'''Ceva's Theorem''' is an algebraic statement regarding the lengths of [[Cevian|cevians]] in a [[triangle]].
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'''Ceva's Theorem''' is a criterion for the [[concurrence]] of [[cevian]]s in a [[triangle]].
  
  
 
== Statement ==
 
== Statement ==
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A [[necessary and sufficient]] condition for <math>AD, BE, CF,</math> where <math>D, E,</math> and <math>F</math> are points of the respective side lines <math>BC, CA, AB</math> of a triangle <math>ABC</math>, to be concurrent is that
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<br><center><math>BD\cdot CE\cdot AF = DC \cdot EA \cdot FB</math></center><br>
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Let <math> \displaystyle ABC </math> be a triangle, and let <math> \displaystyle D, E, F </math> be points on lines <math> \displaystyle BC, CA, AB </math>, respectively.  Lines <math> \displaystyle AD, BE, CF </math> [[concur]] iff if and only if
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where all segments in the formula are [[directed segments]].
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<br><center>
 +
<math>\frac{BD}{DC} \cdot \frac{CE}{EA}\cdot \frac{AF}{FB} = 1 </math>,
 +
</center><br>
 +
where lengths are [[directed segments | directed]].
  
 
[[Image:Ceva1.PNG|center]]
 
[[Image:Ceva1.PNG|center]]
 +
 +
(Note that the cevians do not necessarily lie within the triangle, although they do in this diagram.)
  
 
== Proof ==
 
== Proof ==
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Letting the [[altitude]] from <math>A</math> to <math>BC</math> have length <math>h</math> we have <math>[ABD]=\frac 12 BD\cdot h</math> and <math>[ACD]=\frac 12 DC\cdot h</math> where the brackets represent [[area]].  Thus <math>\frac{[ABD]}{[ACD]} = \frac{BD}{DC}</math>.  In the same manner, we find that <math>\frac{[XBD]}{[XCD]} = \frac{BD}{DC}</math>.  Thus <center><math> \frac{BD}{DC} = \frac{[ABD]}{[ACD]} = \frac{[XBD]}{[XCD]} = \frac{[ABD]-[XBD]}{[ACD]-[XCD]} = \frac{[ABX]}{[ACX]}. </math></center>
 
  
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Likewise, we find that
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We will use the notation <math> \displaystyle [ABC] </math> to denote the area of a triangle with vertices <math> \displaystyle A,B,C </math>.
  
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{| class="wikitable" style="margin: 1em auto 1em auto;height:100px"
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First, suppose <math> \displaystyle AD, BE, CF </math> meet at a point <math> \displaystyle X </math>.  We note that triangles <math> \displaystyle ABD, ADC </math> have the same altitude to line <math> \displaystyle BC </math>, but bases <math> \displaystyle BD </math> and <math> \displaystyle DC </math>.  It follows that <math> \frac {BD}{DC} = \frac{[ABD]}{[ADC]} </math>.  The same is true for triangles <math> \displaystyle XBD, XDC </math>, so
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| <math>\frac{CE}{EA}</math> || <math>=\frac{[BCX]}{[ABX]}</math>
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<center><math> \frac{BD}{DC} = \frac{[ABD]}{[ADC]} = \frac{[XBD]}{[XDC]} = \frac{[ABD]- [XBD]}{[ADC]-[XDC]} = \frac{[ABX]}{[AXC]} </math>. </center>
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|-
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Similarly, <math> \frac{CE}{EA} = \frac{[BCX]}{[BXA]} </math> and <math> \frac{AF}{FB} = \frac{[CAX]}{[CXB]} </math>,
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| <math>\frac{AF}{FB}</math> || <math>=\frac{[ACX]}{[BCX]}</math>
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so
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|}
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<center>
 +
<math> \frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = \frac{[ABX]}{[AXC]} \cdot \frac{[BCX]}{[BXA]} \cdot \frac{[CAX]}{[CXB]} = 1 </math>.
 +
</center>
  
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Thus <center><math> \frac{BD}{DC}\cdot \frac{CE}{EA}\cdot \frac{AF}{FB} = \frac{[ABX]}{[ACX]}\cdot \frac{[BCX]}{[ABX]}\cdot \frac{[ACX]}{[BCX]} = 1  \Rightarrow BD\cdot CE\cdot AF = DC \cdot EA \cdot FB. </math></center>
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Now, suppose <math> \displaystyle D, E,F </math> satisfy Ceva's criterion, and suppose <math> \displaystyle AD, BE </math> intersect at <math> \displaystyle X </math>.  Suppose the line <math> \displaystyle CX </math> intersects line <math> \displaystyle AB </math> at <math> \displaystyle F' </math>.  We have proven that <math> \displaystyle F' </math> must satisfy Ceva's criterion.  This means that <center><math> \frac{AF'}{F'B} = \frac{AF}{FB} </math>, </center> so <center><math> \displaystyle F' = F </math>, </center> and line <math> \displaystyle CF </math> concurrs with <math> \displaystyle AD </math> and <math> \displaystyle BE </math>.  {{Halmos}}
  
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<math>\mathcal{QED}</math>
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== Trigonometric Form ==
  
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== Alternate Formulation ==
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The [[trig | trigonometric]] form of Ceva's Theorem (Trig Ceva) states that cevians <math> \displaystyle AD,BE,CF</math> concur if and only if
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<center>
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The [[trig]] version of Ceva's Theorem states that cevians <math>AD,BE,CF</math> are concurrent if and only if
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<math> \frac{\sin BAD}{\sin DAC} \cdot \frac{\sin CBE}{\sin EBA} \cdot \frac{\sin ACF}{\sin FCB} = 1.</math>
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</center>
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<center><math>\sin BAD \sin ACF \sin CBE = \sin DAC \sin FCB \sin EBA.</math></center>
 
  
 
=== Proof ===
 
=== Proof ===
  
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''This proof is incomplete.  If you can finish it, please do so.  Thanks!''
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First, suppose <math> \displaystyle AD, BE, CF </math> concur at a point <math> \displaystyle X </math>.  We note that
 +
<center>
 +
<math> \frac{[BAX]}{[XAC]} = \frac{ \frac{1}{2}AB \cdot AX \cdot \sin BAX}{ \frac{1}{2}AX \cdot AC \cdot \sin XAC} = \frac{AB \cdot \sin BAD}{AC \cdot \sin DAC} </math>, </center>
 +
and similarly,
 +
<center>
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<math> \frac{[CBX]}{[XBA]} = \frac{AB \cdot \sin BAD}{AC \cdot \sin DAC} ;\; \frac{[ACX]}{[XCB]} = \frac{CA \cdot \sin ACF}{CB \cdot \sin FCB} </math>. </center>
 +
It follows that
 +
<center>
 +
<math> \frac{\sin BAD}{\sin DAC} \cdot \frac{\sin CBE}{\sin EBA} \cdot \frac{\sin ACF}{\sin FCB} = \frac{AB \cdot \sin BAD}{AC \cdot \sin DAC} \cdot \frac{AB \cdot \sin BAD}{AC \cdot \sin DAC} \cdot \frac{CA \cdot \sin ACF}{CB \cdot \sin FCB} </math> <br> <br>  <math> \qquad = \frac{[BAX]}{[XAC]} \cdot \frac{[CBX]}{[XBA]} \cdot \frac{[ACX]}{[XCB]} = 1 </math>.
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</center>
  
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We will use Ceva's Theorem in the form that was already proven to be true.
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The converse follows by an argument almost identical to that used for the first form of Ceva's Theorem. {{Halmos}}
  
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First, we show that if <math>\sin BAD \sin ACF \sin CBE = \sin DAC \sin FCB \sin EBA</math>, holds true then <math>BD\cdot CE\cdot AF = DC \cdot EA \cdot FB</math> which gives that the cevians are concurrent by Ceva's Theorem.  The [[Law of Sines]] tells us that <center><math>\frac{BD}{\sin BAD} = \frac{AB}{\sin ADB} \Leftrightarrow \sin BAD = \frac{BD}{AB\sin ADB}.</math></center>
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== Examples ==
−
 
 
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Likewise, we get
 
  
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{| class="wikitable" style="margin: 1em auto 1em auto"
 
−
|-
 
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| <math>\sin ACF = \frac{AF}{AC\sin CFA}</math>
 
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|-
 
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| <math>\sin CBE = \frac{CE}{BC\sin BEC}</math>
 
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|-
 
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| <math>\sin CAD = \frac{CD}{AC\sin ADC}</math>
 
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|-
 
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| <math>\sin BCF = \frac{BF}{BC\sin BFC}</math>
 
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|-
 
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| <math>\sin  ABE = \frac{AE}{AB\sin AEB}</math>
 
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|}
 
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Thus
 
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{| class="wikitable"
 
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|-
 
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| <math>\sin BAD \sin ACF \sin CBE = \sin CAD \sin BCF \sin ABE</math>
 
−
|-
 
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| <math>\frac{BD}{AB\sin ADB} \cdot \frac{AF}{AC\sin CFA} \cdot \frac{CE}{BC\sin BEC} = \frac{CD}{AC\sin ADC} \cdot \frac{BF}{BC\sin BFC} \cdot \frac{AE}{AB\sin AEB}</math>
 
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|}
 
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== Examples ==
 
 
# Suppose AB, AC, and BC have lengths 13, 14, and 15.  If <math>\frac{AF}{FB} = \frac{2}{5}</math> and <math>\frac{CE}{EA} = \frac{5}{8}</math>.  Find BD and DC.<br> <br>  If <math>BD = x</math> and <math>DC = y</math>, then <math>10x = 40y</math>, and <math>{x + y = 15}</math>.  From this, we find <math>x = 12</math> and <math>y = 3</math>.
 
# Suppose AB, AC, and BC have lengths 13, 14, and 15.  If <math>\frac{AF}{FB} = \frac{2}{5}</math> and <math>\frac{CE}{EA} = \frac{5}{8}</math>.  Find BD and DC.<br> <br>  If <math>BD = x</math> and <math>DC = y</math>, then <math>10x = 40y</math>, and <math>{x + y = 15}</math>.  From this, we find <math>x = 12</math> and <math>y = 3</math>.
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# See the proof of the concurrency of the altitudes of a triangle at the [[orthocenter]].
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# The concurrence of the altitudes of a triangle at the [[orthocenter]] and the concurrence of the perpendicual bisectors of a triangle at the [[circumcenter]] can both be proven by Ceva's Theorem (the latter is a little harder).  Furthermore, the existance of the [[centroid]] can be shown by Ceva, and the existance of the [[incenter]] can be shown using trig Ceva.  However, there are more elegant methods for proving each of these results, and in any case, any result obtained by classic Ceva's Theorem can be proven using ratios of areas.
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# See the proof of the concurrency of the perpendicual bisectors of a triangle at the [[circumcenter]].
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# The existance of [[isotonic conjugate]]s can be shown by classic Ceva, and the existance of [[isogonal conjugate]]s can be shown by trig Ceva.
  
 
== See also ==
 
== See also ==
 
* [[Menelaus' Theorem]]
 
* [[Menelaus' Theorem]]
−
* [[Stewart's Theorem]]
 

Revision as of 21:54, 7 June 2007

Ceva's Theorem is a criterion for the concurrence of cevians in a triangle.


Statement

Let $\displaystyle ABC$ be a triangle, and let $\displaystyle D, E, F$ be points on lines $\displaystyle BC, CA, AB$, respectively. Lines $\displaystyle AD, BE, CF$ concur iff if and only if


$\frac{BD}{DC} \cdot \frac{CE}{EA}\cdot \frac{AF}{FB} = 1$,


where lengths are directed.

Ceva1.PNG

(Note that the cevians do not necessarily lie within the triangle, although they do in this diagram.)

Proof

We will use the notation $\displaystyle [ABC]$ to denote the area of a triangle with vertices $\displaystyle A,B,C$.

First, suppose $\displaystyle AD, BE, CF$ meet at a point $\displaystyle X$. We note that triangles $\displaystyle ABD, ADC$ have the same altitude to line $\displaystyle BC$, but bases $\displaystyle BD$ and $\displaystyle DC$. It follows that $\frac {BD}{DC} = \frac{[ABD]}{[ADC]}$. The same is true for triangles $\displaystyle XBD, XDC$, so

$\frac{BD}{DC} = \frac{[ABD]}{[ADC]} = \frac{[XBD]}{[XDC]} = \frac{[ABD]- [XBD]}{[ADC]-[XDC]} = \frac{[ABX]}{[AXC]}$.

Similarly, $\frac{CE}{EA} = \frac{[BCX]}{[BXA]}$ and $\frac{AF}{FB} = \frac{[CAX]}{[CXB]}$, so

$\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = \frac{[ABX]}{[AXC]} \cdot \frac{[BCX]}{[BXA]} \cdot \frac{[CAX]}{[CXB]} = 1$.

Now, suppose $\displaystyle D, E,F$ satisfy Ceva's criterion, and suppose $\displaystyle AD, BE$ intersect at $\displaystyle X$. Suppose the line $\displaystyle CX$ intersects line $\displaystyle AB$ at $\displaystyle F'$. We have proven that $\displaystyle F'$ must satisfy Ceva's criterion. This means that

$\frac{AF'}{F'B} = \frac{AF}{FB}$,

so

$\displaystyle F' = F$,

and line $\displaystyle CF$ concurrs with $\displaystyle AD$ and $\displaystyle BE$. ∎

Trigonometric Form

The trigonometric form of Ceva's Theorem (Trig Ceva) states that cevians $\displaystyle AD,BE,CF$ concur if and only if

$\frac{\sin BAD}{\sin DAC} \cdot \frac{\sin CBE}{\sin EBA} \cdot \frac{\sin ACF}{\sin FCB} = 1.$

Proof

First, suppose $\displaystyle AD, BE, CF$ concur at a point $\displaystyle X$. We note that

$\frac{[BAX]}{[XAC]} = \frac{ \frac{1}{2}AB \cdot AX \cdot \sin BAX}{ \frac{1}{2}AX \cdot AC \cdot \sin XAC} = \frac{AB \cdot \sin BAD}{AC \cdot \sin DAC}$,

and similarly,

$\frac{[CBX]}{[XBA]} = \frac{AB \cdot \sin BAD}{AC \cdot \sin DAC} ;\; \frac{[ACX]}{[XCB]} = \frac{CA \cdot \sin ACF}{CB \cdot \sin FCB}$.

It follows that

$\frac{\sin BAD}{\sin DAC} \cdot \frac{\sin CBE}{\sin EBA} \cdot \frac{\sin ACF}{\sin FCB} = \frac{AB \cdot \sin BAD}{AC \cdot \sin DAC} \cdot \frac{AB \cdot \sin BAD}{AC \cdot \sin DAC} \cdot \frac{CA \cdot \sin ACF}{CB \cdot \sin FCB}$

$\qquad = \frac{[BAX]}{[XAC]} \cdot \frac{[CBX]}{[XBA]} \cdot \frac{[ACX]}{[XCB]} = 1$.

The converse follows by an argument almost identical to that used for the first form of Ceva's Theorem. ∎

Examples

  1. 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 $y = 3$.
  2. The concurrence of the altitudes of a triangle at the orthocenter and the concurrence of the perpendicual bisectors of a triangle at the circumcenter can both be proven by Ceva's Theorem (the latter is a little harder). Furthermore, the existance of the centroid can be shown by Ceva, and the existance of the incenter can be shown using trig Ceva. However, there are more elegant methods for proving each of these results, and in any case, any result obtained by classic Ceva's Theorem can be proven using ratios of areas.
  3. The existance of isotonic conjugates can be shown by classic Ceva, and the existance of isogonal conjugates can be shown by trig Ceva.

See also

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