EMT 668 Final Project

Summer 1998

by Mark Cowart

A.* Consider any triangle ABC. Select a point P inside
and draw lines AP, BP, and CP extended to their intersections with the opposite
sides D, E, and F respectively. Explore (AF)(BD)(EC) and (FB)(DC)(EA) for
various triangles and locations of P.*

**Click here**
for a GSP sketch of this triangle with a movable Point P. As I moved Point
P around inside triangle ABC the **product** of the measures of certain
segments remained equal. Specifically, (AF)(BD)(EC) is equal to (FB)(DC)(EA)
no matter where Point P is located in the triangle. This implies that the
ratio of

. This conjecture seems to fit the picture and the given facts.

B. *Conjecture? Prove it! Can the results be generalized
(using lines rather than segments to construct triangle ABC) so that point
P can be outside the triangle?*

**According to Ceva's Theorem, in a triangle ABC, three
lines AD, BE, and CF intersect at a single point P if and only if AE/EC
* CD/DB * BF/FA =1**

The proof flows as follows:

**PART 1 In a triangle ABC, if three lines AD, BE, and
CF intersect at a single point P, then AE/EC * CD/DB/ * BF/FA = 1 **

Extend the lines BE and CF beyond the triangle until they meet GH, the line through A parallel to BC. There are several pairs of similar triangles including AHE and BCE, AFG and BCF, AGP and CDP, and finally BDP and AHP. These can be shown similar through alternate interior angles and vertical angles which lead to AA Similarity (see figure below).

From the similar triangles these proportional sides are derived:

1. AE/EC = AH/BC

2. BF/FA = CB/AG

3. AG/CD = AP/DP

4. AH/DB = AP/DP

The last two (#3 and 4) are both linked by AP/DP which yields AG/CD = AH/DB. This proportion is the same as CD/DB = AG/AH. The product of this with the first two gives AE/EC * CD/DB/ * BF/FA = AH/BC*AG/AH*CB/AG which leads to (AH*AG*CB)/(BC*AH*AG) =1. Therefore, if the lines AD, BE, and CF intersect in a single point P, the identity equaling 1 has been proven.

**PART 2 In a triangle ABC, if AE/EC * CD/DB * BF/FA =
1, then the three lines intersect at a single point P.**

Assume that there are at least two points of intersection between the lines.This implies that P and a P' can be the point of intersection of BE and CF and AD; next draw the line AP' until its intersection with BC at a point D' (possible if there are two points of concurrency). Then, from the previously proven part one above, it follows that AE/EC*CD'/D'B*FB/AF=1.

Will a point D' be the same as D?

Also, it has been specifically shown from part one that AE/EC * CD/DB * BF/FA=1. Combining the two yields CD'/D'B=CD/DB or CD'/D'B + 1 = CD/DB +1; these lead to (CD'+D'B)/D'B = (CD+DB)/DB. Finally, by segment addition, CB/D'B = CB/DB which implies that D'B=DB. Thus D' and D have to be the same point which contradicts the assumption that there could be two distinct points of concurrency.

The results of this proven conjecture will also apply to points outside the triangle. To do this, extend lines to construct triangle ABC, and follow the proof above for a point P outside the triangle.

C. *Show that the ratio of the areas of ABC and DEF is
always greater than or equal to 4. When is it equal to 4?*

According to the GSP sketch, the ratio of the triangle ABC to triangle DEF will be 4 when triangle ABC is an equilateral triangle and point P is the same point as the centroid of the triangle.

Otherwise, triangle ABC will always have a ratio greater
than 4 to triangle DEF. **Click here** for
GSP sketch with a movable point P that illustrates this.

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