HamiltonHardison’s Exploration of

Cevva’s Theorem

From Jim Wilson’s website:
Ceva's Theorem. Consider any triangle ABC. Select a point P inside the triangle and draw lines AP, BP, and CP extended to their intersections with the opposite sides in points D, E, and F respectively.

1. Explore (AF)(BD)(EC) and (FB)(DC)(EA) for various triangles and various locations of P.

2. Conjecture? Prove it! (you may need draw some parallel lines to produce some similar triangles, OR, you made need to consider areas of triangles within the figure) Also, it probably helps to consider the ratio

3. In your write-up, after the proof, you might want to discuss how this theorem could be used for proving concurrency of the medians (if P is the centroid), the lines of the altitudes (if P is the orthocenter), the bisectors of the angles (if P is the incenter), or the perpendicular bisectors of the sides (if P is the circumcenter). Concurrency of other special points?

4. Explore a generalization of the result (using lines rather than segments to construct ABC) so that point P can be outside the triangle. Show a working GSP sketch.




By constructing a GSP sketch and exploring measurements, it appears that .


At the advice of Clay Kitchings, I began by drawing a line parallel to BC through A and began looking for similar triangles.  The following pairs of similar triangles appeared as a result of Angle-Angle Similarity due to vertical and alternate-interior angles.  For visualization purposes, the interiors of the desired triangles have been shaded.






From the pairs of similar triangles, we can now investigate several ratios:

From 1:                 From 2:                  From 3:                  From 4: 


By combining ratios 1 and 3 (since ), we have . (ratio 5)

From ratio 2, we have:    (6).  From ratio 4, we have:     (7).

By replacing AI and AH in ratio 5 with the results from equations (6) and (7), we have .

By multiplication, we can transform this equation to .


Since , their ratio is 1.  So our result is .  Or , as desired.


General Case:

Here is a sketch to investigate cases when p is outside of the triangle.

It appears that the theorem holds as long as p does not line on the lines AB, BC, or CB as this will result in an indeterminant expression (which GSP has some difficulty handling as it reports zero, undefined, infinity, or numerical values).  To investigate this case, select point p and one of the lines.  Then, from the edit menu, select merge point to line. 

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