Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Ceva's theorem"

(Created page with "'''Ceva's theorem''' is a criterion for the concurrence of cevians in a triangle. == Statement == thumb|right Let <math>ABC </math> be a tri...")
 
(→‎Proof)
 
(13 intermediate revisions by 7 users not shown)
Line 9: Line 9:
 
<math>\frac{BD}{DC} \cdot \frac{CE}{EA}\cdot \frac{AF}{FB} = 1 </math>,
 
<math>\frac{BD}{DC} \cdot \frac{CE}{EA}\cdot \frac{AF}{FB} = 1 </math>,
 
</center><br>
 
</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>.
+
where lengths are [[directed legnths | directed]]. This also works for the [[reciprocal]] of each of the ratios, as the reciprocal of <math>1</math> is <math>1</math>.
  
  
Line 15: Line 15:
  
  
−
The proof using [[Routh's theorem]] is extremely trivial, so we will not include it.
+
The proof using [[Routh's Theorem]] is extremely trivial, so we will not include it.
  
 
== Proof ==
 
== Proof ==
Line 38: Line 38:
 
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>
 
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:
+
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>
 
<math>xyz=\frac{1-e}{e}\cdot{z}\cdot\frac{1-f}{f}\cdot{x}\cdot\frac{1-d}{d}y</math>
Line 78: Line 78:
  
 
===Intermediate===
 
===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>)
+
*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.
 +
 
 +
*Let <math>M</math> be the midpoint of side <math>AB</math> of triangle <math>ABC</math>. Points <math>D</math> and <math>E</math> lie on line segments <math>BC</math> and <math>CA</math>, respectively, such that <math>DE</math> and <math>AB</math> are parallel. Point <math>P</math> lies on line segment <math>AM</math>. Lines <math>EM</math> and <math>CP</math> intersect at <math>X</math> and lines <math>DP</math> and <math>CM</math> meet at <math>Y</math>. Prove that <math>X,Y,B</math> are collinear. ([[Ceva I.2]])
  
 
== See also ==
 
== See also ==
 
* [[Stewart's theorem]]
 
* [[Stewart's theorem]]
 
* [[Menelaus' theorem]]
 
* [[Menelaus' theorem]]
 +
* [[Routh's Theorem]]
  
 
[[Category:Geometry]]
 
[[Category:Geometry]]
  
 
[[Category:Theorems]]
 
[[Category:Theorems]]

Latest revision as of 14:56, 3 November 2024

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 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, the 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.
  • Let $M$ be the midpoint of side $AB$ of triangle $ABC$. Points $D$ and $E$ lie on line segments $BC$ and $CA$, respectively, such that $DE$ and $AB$ are parallel. Point $P$ lies on line segment $AM$. Lines $EM$ and $CP$ intersect at $X$ and lines $DP$ and $CM$ meet at $Y$. Prove that $X,Y,B$ are collinear. (Ceva I.2)

See also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Ceva%27s_theorem&oldid=231398"