Difference between revisions of "Ceva's Theorem"

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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 ==
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
 
<br><center><math>BD\cdot CE\cdot AF = DC \cdot EA \cdot FB</math></center><br>
 
where all segments in the formula are [[directed segments]].
 
  
[[Image:Ceva1.PNG|center]]
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[[Image:Ceva1.PNG|thumb|right]]
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Let <math>ABC </math> be a triangle, and let <math>D, E, F  </math> be points on lines <math>BC, CA, AB </math>, respectively.  Lines <math>AD, BE, CF </math> are [[concurrent]] if and only if
 +
<br><center>
 +
<math>\frac{BD}{DC} \cdot \frac{CE}{EA}\cdot \frac{AF}{FB} = 1 </math>,
 +
</center><br>
 +
where lengths are [[directed segments | directed]]. This also works for the [[reciprocal]] of each of the ratios, as the reciprocal of <math>1</math> is <math>1</math>.
 +
 
 +
 
 +
(Note that the cevians do not necessarily lie within the triangle, although they do in this diagram.)
 +
 
 +
 
 +
The proof using [[Routh's Theorem]] is extremely trivial, so we will not include it.
  
 
== Proof ==
 
== Proof ==
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>
 
  
Likewise, we find that  
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We will use the notation <math>[ABC] </math> to denote the area of a triangle with vertices <math>A,B,C </math>.
 +
 
 +
First, suppose <math>AD, BE, CF </math> meet at a point <math>X </math>.  We note that triangles <math>ABD, ADC </math> have the same altitude to line <math>BC </math>, but bases <math>BD </math> and <math>DC </math>.  It follows that <math> \frac {BD}{DC} = \frac{[ABD]}{[ADC]} </math>.  The same is true for triangles <math>XBD, XDC </math>, so
 +
 
 +
<center><math> \frac{BD}{DC} = \frac{[ABD]}{[ADC]} = \frac{[XBD]}{[XDC]} = \frac{[ABD]- [XBD]}{[ADC]-[XDC]} = \frac{[ABX]}{[AXC]} </math>. </center>
 +
Similarly, <math> \frac{CE}{EA} = \frac{[BCX]}{[BXA]} </math> and <math> \frac{AF}{FB} = \frac{[CAX]}{[CXB]} </math>,
 +
so
 +
<center>
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<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>
 +
 
 +
Now, suppose <math>D, E,F </math> satisfy Ceva's criterion, and suppose <math>AD, BE </math> intersect at <math>X </math>.  Suppose the line <math>CX </math> intersects line <math>AB </math> at <math>F' </math>.  We have proven that <math>F' </math> must satisfy Ceva's criterion.  This means that <center><math> \frac{AF'}{F'B} = \frac{AF}{FB} </math>, </center> so <center><math>F' = F </math>, </center> and line <math>CF </math> concurs with <math>AD </math> and <math>BE </math>.  {{Halmos}}
  
{| class="wikitable" style="margin: 1em auto 1em auto;height:100px"
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==Proof by [[Barycentric coordinates]]==
| <math>\frac{CE}{EA}</math> || <math>=\frac{[BCX]}{[ABX]}</math>
 
|-
 
| <math>\frac{AF}{FB}</math> || <math>=\frac{[ACX]}{[BCX]}</math>
 
|}
 
  
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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Since <math>D\in BC</math>, we can write its coordinates as <math>(0,d,1-d)</math>. The equation of line <math>AD</math> is then <math>z=\frac{1-d}{d}y</math>.
  
<math>\mathcal{QED}</math>
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Similarly, since <math>E=(1-e,0,e)</math>, and <math>F=(f,1-f,0)</math>, we can see that the equations of <math>BE</math> and <math>CF</math> respectively are <math>x=\frac{1-e}{e}z</math> and <math>y=\frac{1-f}{f}x</math>
  
== Alternate Formulation ==
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[[Multiplying]] the three together yields the solution to the equation:
  
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>xyz=\frac{1-e}{e}\cdot{z}\cdot\frac{1-f}{f}\cdot{x}\cdot\frac{1-d}{d}y</math>
  
<center>$\sin BAD \sin ACF \sin CBE = \sin DAC \sin FCB \sin EBA.$</center>
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Dividing by <math>xyz</math> yields:
  
=== Proof ===
 
  
''This proof is incomplete.  If you can finish it, please do so.  Thanks!''
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<math>1=\frac{1-e}{e}\cdot\frac{1-f}{f}\cdot\frac{1-d}{d}</math>, which is equivalent to Ceva's theorem
 +
 
 +
QED
 +
 
 +
== Trigonometric Form ==
  
We will use Ceva's Theorem in the form that was already proven to be true.
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The [[trig | trigonometric]] form of Ceva's Theorem (Trig Ceva) states that cevians <math>AD,BE,CF</math> concur if and only if
 +
<center>
 +
<math> \frac{\sin BAD}{\sin DAC} \cdot \frac{\sin CBE}{\sin EBA} \cdot \frac{\sin ACF}{\sin FCB} = 1.</math>
 +
</center>
  
First, we show that if $\sin BAD \sin ACF \sin CBE = \sin DAC \sin FCB \sin EBA$, holds true then $BD\cdot CE\cdot AF = DC \cdot EA \cdot FB$ which gives that the cevians are concurrent by Ceva's Theorem.  The [[Law of Sines]] tells us that <center>$\frac{BD}{\sin BAD} = \frac{AB}{\sin ADB} \Leftrightarrow \sin BAD = \frac{BD}{AB\sin ADB}.$</center>
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=== Proof ===
  
Likewise, we get
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First, suppose <math>AD, BE, CF </math> concur at a point <math>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>
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and similarly,
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<center>
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<math> \frac{[CBX]}{[XBA]} = \frac{BC \cdot \sin CBE}{BA \cdot \sin EBA} ;\; \frac{[ACX]}{[XCB]} = \frac{CA \cdot \sin ACF}{CB \cdot \sin FCB} </math>. </center>
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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{BC \cdot \sin CBE}{BA \cdot \sin EBA} \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>
  
{| class="wikitable" style="margin: 1em auto 1em auto"
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Here, sign is irrelevant, as we may interpret the sines of [[directed angles]] mod <math>\pi </math> to be either positive or negative.
|-
 
| $\sin ACF = \frac{AF}{AC\sin CFA}$
 
|-
 
| $\sin CBE = \frac{CE}{BC\sin BEC}$
 
|-
 
| $\sin CAD = \frac{CD}{AC\sin ADC}$
 
|-
 
| $\sin BCF = \frac{BF}{BC\sin BFC}$
 
|-
 
| $\sin  ABE = \frac{AE}{AB\sin AEB}$
 
|}
 
  
Thus
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The converse follows by an argument almost identical to that used for the first form of Ceva's Theorem.  {{Halmos}}
  
{| class="wikitable"
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== Problems ==
|-
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===Introductory===
| <math>\sin BAD \sin ACF \sin CBE = \sin CAD \sin BCF \sin ABE</math>
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*Suppose <math>AB, AC</math>, and <math>BC</math> have lengths <math>13, 14</math>, and <math>15</math>, respectively.  If <math>\frac{AF}{FB} = \frac{2}{5}</math> and <math>\frac{CE}{EA} = \frac{5}{8}</math>,  find <math>BD</math> and <math>DC</math>. ([[Ceva's Theorem/Problems|Source]])
|-
 
| $\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}
 
|}
 
  
== Examples ==
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===Intermediate===
# 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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*In <math>\Delta ABC, AD, BE, CF</math> are concurrent lines. <math>P, Q, R</math> are points on <math>EF, FD, DE</math> such that <math>DP, EQ, FR</math> are concurrent. Prove that (using ''plane geometry'') <math>AP, BQ, CR</math> are concurrent. (<url>viewtopic.php?f=151&t=543574 </url>)
# See the proof of the concurrency of the altitudes of a triangle at the [[orthocenter]].
 
# See the proof of the concurrency of the perpendicual bisectors of a triangle at the [[circumcenter]].
 
  
 
== See also ==
 
== See also ==
 +
* [[Stewart's Theorem]]
 
* [[Menelaus' Theorem]]
 
* [[Menelaus' Theorem]]
* [[Stewart's Theorem]]
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 +
[[Category:Geometry]]
 +
 
 +
[[Category:Theorems]]

Latest revision as of 02:35, 24 July 2020

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


Statement

Ceva1.PNG

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


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


where lengths are directed. This also works for the reciprocal of each of the ratios, as the reciprocal of $1$ is $1$.


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


The proof using Routh's Theorem is extremely trivial, so we will not include it.

Proof

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

First, suppose $AD, BE, CF$ meet at a point $X$. We note that triangles $ABD, ADC$ have the same altitude to line $BC$, but bases $BD$ and $DC$. It follows that $\frac {BD}{DC} = \frac{[ABD]}{[ADC]}$. The same is true for triangles $XBD, XDC$, 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 $D, E,F$ satisfy Ceva's criterion, and suppose $AD, BE$ intersect at $X$. Suppose the line $CX$ intersects line $AB$ at $F'$. We have proven that $F'$ must satisfy Ceva's criterion. This means that

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

so

$F' = F$,

and line $CF$ concurs with $AD$ and $BE$.

Proof by Barycentric coordinates

Since $D\in BC$, we can write its coordinates as $(0,d,1-d)$. The equation of line $AD$ is then $z=\frac{1-d}{d}y$.

Similarly, since $E=(1-e,0,e)$, and $F=(f,1-f,0)$, we can see that the equations of $BE$ and $CF$ respectively are $x=\frac{1-e}{e}z$ and $y=\frac{1-f}{f}x$

Multiplying the three together yields the solution to the equation:

$xyz=\frac{1-e}{e}\cdot{z}\cdot\frac{1-f}{f}\cdot{x}\cdot\frac{1-d}{d}y$

Dividing by $xyz$ yields:


$1=\frac{1-e}{e}\cdot\frac{1-f}{f}\cdot\frac{1-d}{d}$, which is equivalent to Ceva's theorem

QED

Trigonometric Form

The trigonometric form of Ceva's Theorem (Trig Ceva) states that cevians $AD,BE,CF$ 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 $AD, BE, CF$ concur at a point $X$. 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,

$\frac{[CBX]}{[XBA]} = \frac{BC \cdot \sin CBE}{BA \cdot \sin EBA} ;\; \frac{[ACX]}{[XCB]} = \frac{CA \cdot \sin ACF}{CB \cdot \sin FCB}$.

It follows that

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

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

Here, sign is irrelevant, as we may interpret the sines of directed angles mod $\pi$ to be either positive or negative.

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

Problems

Introductory

  • Suppose $AB, AC$, and $BC$ have lengths $13, 14$, and $15$, respectively. If $\frac{AF}{FB} = \frac{2}{5}$ and $\frac{CE}{EA} = \frac{5}{8}$, find $BD$ and $DC$. (Source)

Intermediate

  • In $\Delta ABC, AD, BE, CF$ are concurrent lines. $P, Q, R$ are points on $EF, FD, DE$ such that $DP, EQ, FR$ are concurrent. Prove that (using plane geometry) $AP, BQ, CR$ are concurrent. (<url>viewtopic.php?f=151&t=543574 </url>)

See also

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