EMT 668
Final Project #4
Exploration of any triangle ABC and a point P
by
Brian Seitz

This exploration is accomplished using Geometer's Sketchpad. To begin I constructed a triangle ABC (yellow shaded area) and a random point P inside the triangle ABC. Next, I constructed 3 unique lines, one from each vertex through point P. Points D, E , and F are the points where the lines intersected triangle ABC. Points DEF were connected to form a triangle (green shaded area).

Exploring segments AF, BD, and CE, compared to segments FB, DC, and EA, I could made interesting hypotheses. The ratio was found to equal precisely 1.00. No matter where point P was located inside triangle ABC this ratio of (AF)(BD)(CE)/(FB)(DC)(EA) equaled 1.00.

Also, I compared the area of triangle ABC to the area of triangle DEF. The ratio of ABC/DEF never got below 4.00.


To explore the above construction, click here

Proving that the ratio is always equal to 1.00 is done by drawing parallel lines to line CP through points A and B, respectively. From this we can now prove that (AF)(BD)(CE)/(FB)(DC)(EA)=
1.00 or (AF)(BD)(CE) = (FB)(DC)(EA).



PROOF:
Triangle PBF is similar to triangle IBA and triangle PAF is similar to triangle JAB, because corresponding angles are congruent. Now, we get

and since triangle AEI is similar to triangle CEP and triangle BDJ is similar to CDP ( alternate interior angles and vertical angles are congruent), we can say that

From the similar triangles we can get the following ratios:

,

Working with the second two ratios I got,

Then, using the first two ratios, along with this new one, I got

Then simplifying the right side of the equation gives me

Constructing triangle ABC again, but this time using lines instead of segments, and placing point P outside the triangle gives me a picture as such. One can see that the ratio remains 1.00. The ratio will remain 1.00 for wherever the point P is placed. To see this construction on GSP, click here.


To find out when the ratio of the area of triangle ABC to the area of triangle DEF equals 4.00, I hypothesized that this would occur when all four triangles inside triangle ABC were equal, regarding their area. So, triangle DEF, triangle AEF. triangle BFD, and triangle CED must all have equal areas. The most logical way this would occur would to have equal sides involved and I could make the sides equal by constructing the midpoints of segments AB, BC, and AC, then constructing the three medians of triangle ABC. As shown below, this gives me four equal triangles inside triangle ABC, thus the ratio of triangle ABC to triangle DEF = 4.00. To see this construction on GSP, click here


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