View source for Ceva's Theorem

'''Ceva's Theorem''' is a criterion for the [[concurrence]] of [[cevian]]s in a [[triangle]].

== Statement ==

[[Image:Ceva1.PNG|thumb|right]]
Let <math>\triangle 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
<cmath>\frac{BD}{DC} \cdot \frac{CE}{EA}\cdot \frac{AF}{FB} = 1</cmath>
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>.

(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 <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 
<cmath>\frac{BD}{DC} = \frac{[ABD]}{[ADC]} = \frac{[XBD]}{[XDC]} = \frac{[ABD]- [XBD]}{[ADC]-[XDC]} = \frac{[ABX]}{[AXC]}</cmath>
(Note the result is just the area lemma) Similarly, <math>\frac{CE}{EA} = \frac{[BCX]}{[BXA]}</math> and <math>\frac{AF}{FB} = \frac{[CAX]}{[CXB]}</math>, so
<cmath>\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = \frac{[ABX]}{[AXC]} \cdot \frac{[BCX]}{[BXA]} \cdot \frac{[CAX]}{[CXB]} = 1</cmath>
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 <cmath>\frac{AF'}{F'B} = \frac{AF}{FB}</cmath> so <cmath>F' = F</cmath> and line <math>CF</math> concurs with <math>AD</math> and <math>BE</math>. <math>\square</math>

== 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:
<cmath>xyz=\frac{1-e}{e}\cdot{z}\cdot\frac{1-f}{f}\cdot{x}\cdot\frac{1-d}{d}y</cmath>

Dividing by <math>xyz</math> yields:
<cmath>1=\frac{1-e}{e}\cdot\frac{1-f}{f}\cdot\frac{1-d}{d}</cmath>, which is equivalent to Ceva's Theorem.

<cmath>Q.E.D.</cmath>

== Trigonometric Form ==

The [[trigonometry|trigonometric]] form of Ceva's Theorem states that cevians <math>AD,BE,CF</math> concur if and only if
<cmath>\frac{\sin BAD}{\sin DAC} \cdot \frac{\sin CBE}{\sin EBA} \cdot \frac{\sin ACF}{\sin FCB} = 1</cmath>

=== Proof ===

First, suppose <math>AD, BE, CF </math> concur at a point <math>X </math>.  We note that
<cmath>\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}</cmath>
and similarly,
<cmath>\frac{[CBX]}{[XBA]} = \frac{BC \cdot \sin CBE}{BA \cdot \sin EBA} ;\; \frac{[ACX]}{[XCB]} = \frac{CA \cdot \sin ACF}{CB \cdot \sin FCB}</cmath>
It follows that
<cmath>\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}  </cmath> <br> <cmath>\qquad = \frac{[BAX]}{[XAC]} \cdot \frac{[CBX]}{[XBA]} \cdot \frac{[ACX]}{[XCB]} = 1</cmath>

Here, the 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. <math>\square</math>

== 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.

*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's Theorem/Problems|Source]])

== See Also ==

* [[Stewart's Theorem]]
* [[Menelaus' Theorem]]
* [[Routh's Theorem]]

[[Category:Geometry]]
[[Category:Theorems]]