Ceva's Theorem

Revision as of 11:03, 18 August 2006 by Joml88 (talk | contribs)

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


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

$BD\cdot CE\cdot AF = DC \cdot EA \cdot FB$

where all segments in the formula are directed segments.



Letting the altitude from $A$ to $BC$ have length $h$ we have $[ABD]=\frac 12 BD\cdot h$ and $[ACD]=\frac 12 DC\cdot h$ where the brackets represent area. Thus $\frac{[ABD]}{[ACD]} = \frac{BD}{DC}$. In the same manner, we find that $\frac{[XBD]}{[XCD]} = \frac{BD}{DC}$. Thus

$\frac{BD}{DC} = \frac{[ABD]}{[ACD]} = \frac{[XBD]}{[XCD]} = \frac{[ABD]-[XBD]}{[ACD]-[XCD]} = \frac{[ABX]}{[ACX]}.$

Likewise, we find that

$\frac{CE}{EA}$ $=\frac{[BCX]}{[ABX]}$
$\frac{AF}{FB}$ $=\frac{[ACX]}{[BCX]}$


$\frac{BD}{DC}\cdot \frac{CE}{EA}\cdot \frac{AF}{FB} = \frac{[ABX]}{[ACX]}\cdot \frac{[BCX]}{[ABX]}\cdot \frac{[ACX]}{[BCX]} = 1  \Rightarrow BD\cdot CE\cdot AF = DC \cdot EA \cdot FB.$



  1. 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 $y = 3$.
  2. See the proof of the concurrency of the altitudes of a triangle at the orthocenter.
  3. See the proof of the concurrency of the perpendicual bisectors of a triangle at the circumcenter.

See also

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