Final Project

By: Laura Lowe

Problem 2: 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.

2. I found from my investigations with GSP that (AF)(BD)(CE) = (BF)(CD)(AE) no matter how I manipulated the triangle. In order to prove (AF)(BD)(CE) = (BF)(CD)(AE), I am going to show that their ratio is 1.

Proof:

Construct a line that passes through A and is parallel to BC. This line intersects with BE and CF, label these points R and S. (Figure 1)

Figure 1

This construction gave me several sets of similar triangles.

rAPS is similar to rCPD since LAPS LCPD (vertical angles), and LCDP LPAS (alternate interior angles). (Figure 2)

Figure 2

From which we can conclude: Equation 1

Similarly, rRAP is similar to rDPB (Figure 3),

Figure 3

From which we can conclude: Equation 2

rSAF is similar to rCFB (Figure 4),

Figure 4

From which we can conclude: Equation 3

and, rRAE is similar to rCEB (Figure 5),

Figure 5

From which we can conclude: Equation 4

From Equations 1 and 2

and

use substitution and rearrange to get Equation 5.

From Equations 3, and 4

and

solve for BC

and

Use substitution and rearrange

Divide both sides by AS

Substitute Equation 5

Multiply both sides by CD

Q.E.D.

3. The first proof led me to ask if it can be proved the other way. In other words, given rABC, with a point D on segment BC, a point E on segment AC, and a point F on segment AB, such that,

is it true that AD, BE, and CF must be concurrent?

Proof

Suppose segment AD and BE intersect at a point P. Construct the line CP. Construct the point of intersection with segment AB and call it L. I want to prove that L = D.

By my construction and what I have already proven,

From my given, I have

When I substitute, I get

Therefore D = L. Q.E.D.

From this result, the centroid, orthocenter, and incenter can all be easily proven to be concurrent using Ceva’s Theorem as they are all constructed using the vertices of the triangle. This is true even when the special point is outside the triangle. (See part 4) The only time Ceva’s Theorem can be used to prove concurrency of the circumcenter is for the equilateral triangle as that is the only triangle whose perpendicular bisectors intersect the vertices.

4. Can P lie outside rABC?

I constructed rABC using lines instead of segments and tested Ceva’s Theorem with P outside. Click here to see my construction.

It appears from my construction that Ceva’s Theorem holds for all triangles. Click here to try it in GSP.

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