This write-up is for the Final.

EMAT 6680 Final Project, Fall, 2000

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Part A

Consider any triangle ABC. We will 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.

We will now explore the product (AF)(BD)(EC) and the product (FB)(DC)(EA) for various triangles and various locations of P.

If we measure the segments and do the calculation of the products to be explored we see that the products are equal. The image below shows a different triangle with P in different location.

Once again the two products being explored are equal.

Part B

If we conjecture that the product of (AF)(BD)(EC) will always equal the product of (FB)(DC)(EA) in regard to the conditions stated for triangle ABC in part A, then we can also conjecture that the ratio of these products is equal to 1.

Since no amount of measurement is proof in itself, we will attempt to prove the ratio shown above is equal to 1.

Given: triangle ABC and a point P inside (same conditions in our exploration above),

line UC and line TA are parallel to line BE.

Prove:

1. triangle AUC is similar to triangle APE ...( Angle, Angle Similarity)

• angle CAU is congruent to angle EAP ...(Reflexive Property)
• angle AUC is congruent to angle APE ...(Corresponding angles)

2. AC / AE = UC / PE ...(Corresponding sides of similar triangles are proportional)

3. Therefore, (AC)(PE) = (AE)(UC)

4. triangle TCA is similar to triangle PCE ...(Angle, Angle Similarity)

• angle TCA is congruent to angle PCE ...(Reflexive Property)
• angle CTA is congruent to angle CPE ...(Corresponding angles)

5. AC / CE = TA / PE ...(Corresponding sides of similar triangles are proportional)

6. Therefore, (AC)(PE) = (TA)(CE)

7. triangle FPB is similar to triangle FTA ...(Angle, Angle Similarity)

• angle BFP is congruent angle AFT ...(Vertical angles)
• angle FBP is congruent to angle FAT ...(Alternate interior angles)

8. BF / AF = BP / TA ...(Corresponding sides of similar triangles are proportional)

9. Therefore, TA = [(AF)(BP)] / BF

10. triangle DUC is similar to triangle DPB ...(Angle, Angle Similarity)

• angle CDU is congruent to angle BDP ...(Vertical angles)
• angle UCD is congruent to angle PBD ...(Alternate interior angles)

11. UC / BP = CD / BD ...(Corresponding sides of similar triangles are proportional)

12. Therefore, UC = [(BP)(CD)] / BD

13. (AE)(UC) = (TA)(CE) ...(Substitution ... step 3 and step 6)

14. (AE)[(BP)(CD)] / BD = (CE)[(AF)(BP)] / BF ...(Substitution ... step 13, step 9, and step 12)

15. Hence, (AF)(CE)(BP)(BD) = (BF)(AE)(BP)(CD)

16. Therefore,

If we construct triangle ABC with lines rather than segments, then we may be able to generalize our results from above so that point P can be outside triangle ABC.

Part C

Now we will construct triangle DEF. If point P is inside triangle ABC, then the ratio of the area of triangle ABC to the area of triangle DEF is always greater than or equal to 4.

Furthermore, if the point P is at the centroid (G), then the ratio of the areas will be equal to 4.