Difference between revisions of "Ceva's Theorem"

(Redirected page to Ceva's theorem)
(Tag: New redirect)
 
(35 intermediate revisions by 18 users not shown)
Line 1: Line 1:
'''Ceva's Theorem''' is an algebraic statement regarding the lengths of [[Cevian|cevians]] in a [[triangle]].
+
#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>
 
 
 
 
 
== 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

Redirect to: