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'''Ceva's Theorem''' is an algebraic statement regarding the lengths of [[Cevian|cevians]] in a [[triangle]].
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#REDIRECT[[Ceva's theorem]]
 
 
 
 
== Statement ==
 
A [[necessary and sufficient]] condition for <math>AD, BE, CF,</math> where <math>D, E,</math> and <math>F</math> are points of the respective side lines <math>BC, CA, AB</math> of a triangle <math>ABC</math>, to be concurrent is that
 
<br><center><math>BD\cdot CE\cdot AF = DC \cdot EA \cdot FB</math></center><br>
 
where all segments in the formula are [[directed segments]].
 
 
 
[[Image:Ceva1.PNG|center]]
 
 
 
== Proof ==
 
Letting the [[altitude]] from <math>A</math> to <math>BC</math> have length <math>h</math> we have <math>[ABD]=\frac 12 BD\cdot h</math> and <math>[ACD]=\frac 12 DC\cdot h</math> where the brackets represent [[area]].  Thus <math>\frac{[ABD]}{[ACD]} = \frac{BD}{DC}</math>.  In the same manner, we find that <math>\frac{[XBD]}{[XCD]} = \frac{BD}{DC}</math>.  Thus <center><math> \frac{BD}{DC} = \frac{[ABD]}{[ACD]} = \frac{[XBD]}{[XCD]} = \frac{[ABD]-[XBD]}{[ACD]-[XCD]} = \frac{[ABX]}{[ACX]}. </math></center>
 
 
 
Likewise, we find that
 
 
 
{| class="wikitable" style="margin: 1em auto 1em auto;height:100px"
 
| <math>\frac{CE}{EA}</math> || <math>=\frac{[BCX]}{[ABX]}</math>
 
|-
 
| <math>\frac{AF}{FB}</math> || <math>=\frac{[ACX]}{[BCX]}</math>
 
|}
 
 
 
Thus <center><math> \frac{BD}{DC}\cdot \frac{CE}{EA}\cdot \frac{AF}{FB} = \frac{[ABX]}{[ACX]}\cdot \frac{[BCX]}{[ABX]}\cdot \frac{[ACX]}{[BCX]} = 1  \Rightarrow BD\cdot CE\cdot AF = DC \cdot EA \cdot FB. </math></center>
 
 
 
<math>\mathcal{QED}</math>
 
 
 
== Alternate Formulation ==
 
 
 
The [[trig]] version of Ceva's Theorem states that cevians <math>AD,BE,CF</math> are concurrent if and only if
 
 
 
<center>$\sin BAD \sin ACF \sin CBE = \sin DAC \sin FCB \sin EBA.$</center>
 
 
 
=== Proof ===
 
 
 
''This proof is incomplete.  If you can finish it, please do so.  Thanks!''
 
 
 
We will use Ceva's Theorem in the form that was already proven to be true.
 
 
 
First, we show that if $\sin BAD \sin ACF \sin CBE = \sin DAC \sin FCB \sin EBA$, holds true then $BD\cdot CE\cdot AF = DC \cdot EA \cdot FB$ which gives that the cevians are concurrent by Ceva's Theorem.  The [[Law of Sines]] tells us that <center>$\frac{BD}{\sin BAD} = \frac{AB}{\sin ADB} \Leftrightarrow \sin BAD = \frac{BD}{AB\sin ADB}.$</center>
 
 
 
Likewise, we get
 
 
 
{| class="wikitable" style="margin: 1em auto 1em auto"
 
|-
 
| $\sin ACF = \frac{AF}{AC\sin CFA}$
 
|-
 
| $\sin CBE = \frac{CE}{BC\sin BEC}$
 
|-
 
| $\sin CAD = \frac{CD}{AC\sin ADC}$
 
|-
 
| $\sin BCF = \frac{BF}{BC\sin BFC}$
 
|-
 
| $\sin  ABE = \frac{AE}{AB\sin AEB}$
 
|}
 
 
 
Thus
 
 
 
{| class="wikitable"
 
|-
 
| <math>\sin BAD \sin ACF \sin CBE = \sin CAD \sin BCF \sin ABE</math>
 
|-
 
| $\frac{BD}{AB\sin ADB} \cdot \frac{AF}{AC\sin CFA} \cdot \frac{CE}{BC\sin BEC} = \frac{CD}{AC\sin ADC} \cdot \frac{BF}{BC\sin BFC} \cdot \frac{AE}{AB\sin AEB}
 
|}
 
 
 
== Examples ==
 
# 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>  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>.
 
# See the proof of the concurrency of the altitudes of a triangle at the [[orthocenter]].
 
# See the proof of the concurrency of the perpendicual bisectors of a triangle at the [[circumcenter]].
 
 
 
== See also ==
 
* [[Menelaus' Theorem]]
 
* [[Stewart's Theorem]]
 

Latest revision as of 16:06, 9 May 2021

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