Ceva's Theorem: For three points D , E and F on \overline{BC} , \overline{CA} and \overline{АB} respectively, we have:

the segments \overline{AD} , \overline{BE} and \overline{CF} have a common intersection point P

if and only if

\frac{|\overline{BD}|}{| \overline{DC}|} \cdot \frac{|\overline{CE}|}{|\overline{EA}|} \cdot \frac{|\overline{AF}|}{|\overline{FB}|} =1

Corollary: The medians of a triangle satisfy Ceva's Theorem and meet at the centroid.


  • Click and drag either of the points D or E, and see how they and the formula in Ceva's Theorem determine the point F.
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  • As always you can click and drag any of the vertices to change the triangle.

The Algorithm in the Interactivity

  • D and E are free (movable) points on \overline{BC} and \overline{CA} , respectively.
  • Define variable NUMerator = |\overline{BD}| \cdot \overline{CE}.
  • Define variable DENominator = |\overline{DC}| \cdot |\overline{EA}| .
  • Define variable Radius = \frac{|\overline{AB}| \cdot NUM}{(NUM+DEN)} .
  • Draw a circle of radius R and centar B.
  • The point F is the intersection of this circle and \overline{AB} .


Definition: A cevian is a line segment joining a vertex (of a triangle) with a point on the opposite side (or its extension).


  • A median is a cevian.
  • The principles of Ceva's Theorem are cevians.

Note The theorem given above is actually a restricted version to cevians on interior points. Here is the statement and proof of Ceva's theorem for all cevians.

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Page last modified on January 30, 2008, at 10:51 AM