Department of Mathematics Education
J. Wilson, EMAT 6680
EMAT 6680
Final Assignment
Matthew Tanner
EMAT 6680
Final Assignment
July 26, 2004
A.
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.
Explore (AF)(BD)(EC) and (FB)(DC)(EA) for various triangles and
various locations of P.
The ratio
of (AF)(BD)(EC) to (FB)(DC)(EA) is one for a number of configurations of
shape and locations of P. The same is true for
polygons of five sides – maybe for all polygons with an odd number of sides. 
B. Conjecture?
Prove it! (you may need draw some parallel lines to produce some similar
triangles). Can the result be generalized (using lines rather than segments
to construct ABC) so that point P can be outside the triangle? Show a working
GSP sketch. There exists a fundamental relationship
of ratios of the segments of sides of a triangle divided by lines emanating
from the opposite vertices and passing through a common central point.

Introducing a line through a vertex
parallel to the opposite side produces useful congruencies. Since
segments WC and AB cross parallel lines and since are opposite angles, they are congruent. So triangles AFW and BFC
are congruent and (1) . Similarly
triangles AEV and BEC are congruent so (2) . Next, observe that
triangles WAP and CDP are congruent and triangles VAP and BDP are congruent. 
From this we can derive (3) . From (1), (2), and (3) we derive

C. Show that when P is inside
triangle ABC, the ratio of the areas of triangle ABC and triangle DEF is
always greater than or equal to 4. When is it equal to 4?. I assert that the ratio of
triangle ABC to triangle DEF is exactly four when the vertices of triangle
DEF are located at the midpoints of the sides of triangle ABC – that is, when
the junction is located at the centroid.
Notice that points E and F each inherit half the horizontal components
of segments AB and AC respectively – each in opposing directions, so segment
EF is half the length of BC.
Similarly segment ED is half that of AB and DF half that of AC. So ED = AF = BF, BD = CD = EF, and DF = CE
= AE. So triangles AEF, BDF, CDE, and
DEF are all equivalent so the ratio of triangle ABC to DEF is exactly 4. Perhaps anecdotally, we
see that when moving the center of the inscribed triangle away from the
centroid, an amount is added to the altitude of the triangle while the base
is lessoned by a scalar multiple resulting in a smaller area and a higher
ratio in comparison to the principle triangle – at least intimating that the
minimum ratio of the principle triangle to the inscribed triangle is 4. 