!FINAL ASSIGNMENT!!!!

BY CAROLYN JOHNSON

UNIVERSITY OF GEORGIA

PROBLEM: Consider any triangle ABC. Select a point P inside the triangle. Draw lines AP, BP, and CP, extending to their intersections with the opposite sides at points D, E, and F respectively.

Let's first look at the graph to see what we are to work with.

Now we are to explore (AF)(BD)(EC), that is, the multiplication of these different line segments,divided by (FB)(DC)(EA) for various triangles and various locations of P. Afterwards we are to come up with a conjecture about what is happening.

First I looked at the graph with point P and the segments, with the values of the segments being shown on the graph. Let's see what I was able to get.

For this location of P, the ratio of the two sets of segment-multiplication is equal to one. Let's relocate P and see what happens with our ratio.

Our ratio is still equal to one even with P at a new location. Let's try another location.

This was a rather drastic move of P. It may be hard to see, but it is located at the B vertex of the triangle ABC. How did it affect our ratio? It didn't change. So let's go outside of the triangle and see if the ratio is changed.

Guess what. The ratio is STILL one. So my conjecture would be that the ratio of the noted line segments will always be equal to one. Do you agree? Let's experiment with our GSP graph to see for ourselves. CLICK HERE FOR GRAPH

PROOF: This is the graph that I will refer to while working my proof. Unfortunately, the graph can't be used as the proof, but it can help us as we do our work.

With this proof I want to prove that (AF)(BD)(CE)/(BF)(CD)(AE) = 1.

Consider triangle ABC with the constructed point P, line CFG, line BEH, and line APD. Construct line GAH to be parallel to BC, while passing through point A. Examining triangle AHE and triangle CBE, we see that angle AHE is equal to angle EBC, by alternating interior angles. Angle ECB is equal to angle AEH, by the same reasoning. Therefore triangle AHE is similar to triangle CBE by angle-angle similarity. With this similarity, we have the ratio AE/EC = AH/BC.

Now let's examine triangles AGF and BCF. Using the parallel lines GAH and BDC, we have angle FGA equal to angleFCB, by alternating interior angles. Angle FBC is equal to angle GAF by the same reasoning. Therefore triangle FGA is similar to triangle FCB by angle-angle similarity. Also, we have the ratio BF/FA = BC/AG.

Comparing triangles AHP and BDP, we see they are similar by angle-angle similarity with the angle PDB being equal to angle PAH and angle PBD equal to angle AHP. The resulting ratio is AG/CD = AP/PD.

For our last pair of similar triangles, we look at triangles AGP and CDP. The angles used to show similarity are angles PCD equal to PGA and GAP equal to PDC. The ratio is AH/BD = AP/PD.

Notice AP/PD = AG/CD and AP/PD = AH/BD. Therefore AG/CD = AH/BD, since they both equal to AP/PD. Mulitplying both sides by CD/AH, we have AG/AH = CD/BD.

With all the above ratios we get: (AF)(BD)(CE)/(BF)(CD)(AE) = 1, which is what I wanted to prove.

To generalize the above result so that it would also hold when point P is located outside the triangle, we would have to use "lines" instead of "segments" in the construction of our triangle ( which is what we did to begin with).

(C) Now we are to show that when P is inside triangle ABC, the ratio of the areas of triangle ABC and triangle DEF is greater than or equal to four.

First we see that the areas of the four smaller triangles are equally the same in the special case where DEF's vertices are located at the midpoints of triangle ABC's sides

With DEF's vertices at ABC's midpoints, a "MEDIAL" triangle is created. The properties of a medial triangle are that (1) it is similar to the original triangle and (2) its area is one-fourth the area of the original triangle. Another way to view the 1: 4 ratio is using the "midpoint theorem." In this case we're examining, the midpoint theorem establishs the congruency of the medial triangle and the original triangle. Therefore the sides of the two triangles are in the ratio of 1: 2. With the "sides" ratio of 1: 2 between the two triangles, the area ratio between the two triangles is 1: 4. This is gotten by squaring the "sides" ratio: (1: 2)^2 = 1/4 = 1: 4. If you would like to test the 1:4 ratio of the medial triangle's area to the original triangle's area

You can see that no matter how you move the vertices around, the ratio of the center triangle to the others will remain constant at .25 (as shown on the graph).

So what happens when we do not have a medial triangle to deal with? Let's look at an example to see what may happen.

This time the ratio shown with the graph represents the ratio of ACB to the smaller FDE. As you experiment with the GSP graph (click below), you will see that the ratio number will not go below 4. This shows that the minimum ratio is 4, but that the ratio can be greater, also. As stated at the beginning, the ratio is greater than or equal to four.