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