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"
(→‎Partial proof by Barycentric Coordinates)
Line 28: Line 28:
 
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}}
 
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}}
  
−
==Partial proof by Barycentric Coordinates==
+
==Proof by Barycentric Coordinates==
  
−
Let three said cevians intersect at point <math>O</math>. Let the barycentric coordinates of <math>O</math> be <math>(X,Y,Z)</math>. Then, because the coordinates are [[homogeneous]], we can say <math>AE=AF=X</math>, <math>BF=BD=Y</math>, and <math>CD=CE=Z</math>. Then, plugging the values into the equation yields:
+
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>\frac{Y}{Z} \cdot \frac{Z}{X}\cdot \frac{X}{Y} = 1 \shadedbox</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>
 +
 
 +
Since the coordinates are [[homogeneous]], we can resize <math>x,y,</math>and<math>z</math> to have <math>xyz=1</math>. This 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 ==
 
== Trigonometric Form ==

Revision as of 21:38, 22 May 2013

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 $AD, BE, CF$ concur iff (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 or 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.)

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:

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

Since the coordinates are homogeneous, we can resize $x,y,$and$z$ to have $xyz=1$. This 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

Olympiad

Other Notes

  • The concurrence of the altitudes of a triangle at the orthocenter and the concurrence of the perpendicular bisectors of a triangle at the circumcenter can both be proven by Ceva's Theorem (the latter is a little harder). Furthermore, the existence of the centroid can be shown by Ceva, and the existence 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 Ceva's Theorem can be obtained using ratios of areas.
  • The existence of isotomic conjugates can be shown by classic Ceva, and the existence of isogonal conjugates can be shown by trig Ceva.

See also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Ceva%27s_Theorem&oldid=52822"