Difference between revisions of "Ceva's Theorem"

(Example)
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== Example ==
 
== Example ==
Suppose AB, AC, and BC have lengths 13, 14, and 15.  If AF:FB = 2:5 and CE:EA = 5:8.  If BD = x and DC = y, then 10x = 40y, and x + y = 15.  From this, we find x = 12 and y = 3.
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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>
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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 also ==
 
== See also ==
 
* [[Menelaus' Theorem]]
 
* [[Menelaus' Theorem]]
 
* [[Stewart's Theorem]]
 
* [[Stewart's Theorem]]

Revision as of 16:10, 20 June 2006

Ceva's Theorem is an algebraic statement regarding the lengths of cevians in a triangle.


Statement

(awaiting image)
A necessary and sufficient condition for AD, BE, CF, where D, E, and F are points of the respective side lines BC, CA, AB of a triangle ABC, to be concurrent is that


$BD * CE * AF = +DC * EA * FB$


where all segments in the formula are directed segments.

Example

Suppose AB, AC, and BC have lengths 13, 14, and 15. If $\frac{AF}{FB} = \frac{2}{5}$ and $\frac{CE}{EA} = \frac{5}{8}$. Find BD and DC.

If $BD = x$ and $DC = y$, then $10x = 40y$, and ${x + y = 15}$. From this, we find $x = 12$ and $y = 3$.

See also

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