Difference between revisions of "Ceva's Theorem"

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'''Ceva's Theorem''' is a criterion for the [[concurrence]] of [[cevian]]s in a [[triangle]].
#REDIRECT[[Ceva's theorem]]
== Statement ==
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
<math>\frac{BD}{DC} \cdot \frac{CE}{EA}\cdot \frac{AF}{FB} = 1 </math>,
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 ==
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>,
<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>.
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}}
==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:
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
== 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
<math> \frac{\sin BAD}{\sin DAC} \cdot \frac{\sin CBE}{\sin EBA} \cdot \frac{\sin ACF}{\sin FCB} = 1.</math>
=== Proof ===
First, suppose <math>AD, BE, CF </math> concur at a point <math>X </math>.  We note that
<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,
<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
<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>.
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 ==
*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]])
*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 ==
* [[Stewart's Theorem]]
* [[Menelaus' Theorem]]

Latest revision as of 16:06, 9 May 2021

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