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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 ==
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''(awaiting image)''<br>
+
 
−
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
+
[[Image:Ceva1.PNG|thumb|right]]
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<br><center><math>BD * CE * AF = +DC * EA * FB</math></center><br>
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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
−
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]]. 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 ==
−
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>
 
−
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:
 
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<center><math>\frac{BX}{XC}=\frac{AB}{CA'},\qquad\frac{CY}{YA}=\frac{CB'}{BA}</math></center>
 
−
<br>
 
−
and thus
 
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<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>.
 
−
<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
 
−
<center><math>\frac{CP}{ZP}=\frac{A'C}{AZ},\qquad\frac{CP}{ZP}=\frac{CB'}{ZB}</math></center>
 
−
<br>
 
−
which lead to
 
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<center><math>\frac{AZ}{ZB}=\frac{A'C}{CB'}</math>.</center>
 
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<br>
 
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Multiplying the last expression with (1) gives
 
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<center><math>\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1</math></center>
 
−
<br>
 
−
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
 
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<center><math>\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1.</math></center>
 
−
<br>
 
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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
 
−
<center><math>\frac{AZ'}{Z'B}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1</math></center>
 
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<br>
 
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and thus
 
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<center><math>\frac{AZ'}{Z'B}=\frac{AZ}{ZB}</math></center>
 
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<br>
 
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which implies <math>Z=Z'</math>, and therefore <math>AX,BY,CZ</math> are concurrent.
 
  
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== Example ==
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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>.
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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>
+
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
−
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>.
+
 
 +
<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>
 +
<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> concurrs with <math>AD </math> and <math>BE </math>.  {{Halmos}}
 +
 
 +
==Proof by [[Barycentric coordinates]]==
 +
 
 +
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>.
 +
 
 +
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>
 +
 
 +
[[Multiplying]] the three together yields the solution to the equation:
 +
 
 +
<math>xyz=\frac{1-e}{e}\cdot{z}\cdot\frac{1-f}{f}\cdot{x}\cdot\frac{1-d}{d}y</math>
 +
 
 +
Dividing by <math>xyz</math> yields:
 +
 
 +
 
 +
<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 ==
 +
 
 +
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>
 +
 
 +
=== Proof ===
 +
 
 +
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>
 +
and similarly,
 +
<center>
 +
<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>
 +
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>.
 +
</center>
 +
 
 +
Here, sign is irrelevant, as we may interpret the sines of [[directed angles]] mod <math>\pi </math> 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.  {{Halmos}}
 +
 
 +
== Problems ==
 +
===Introductory===
 +
*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]])
 +
 
 +
===Intermediate===
 +
*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 also ==
 
== See also ==
 +
* [[Stewart's Theorem]]
 
* [[Menelaus' Theorem]]
 
* [[Menelaus' Theorem]]
−
* [[Stewart's Theorem]]
+
 
 +
[[Category:Geometry]]
 +
 
 +
[[Category:Theorems]]

Revision as of 18:15, 6 August 2018

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 $CF$ concurrs 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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