**Final Project**

** **

**By: Laura Lowe**

**Problem 2: CevaÕs 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 *L*APS *L*CPD (vertical angles), and *L*CDP *L*PAS (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
CevaÕs 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 CevaÕs 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 CevaÕs
Theorem with P outside. Click here
to see my construction.

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

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