**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.**

**B. Conjecture?
Prove it. Consider the ratio**

**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?**

**Exploration
of (AF)(BD)(EC) and (FB)(DC)(EA) for this triangle ABC. Notice
how the product of each of these three segments is equal to 70.34
cm^2.**

**I decided
to take a variety of measurements of the triangles and sides of
this particular triangle ABC.**

**I found that
the ratio of AF to FB = 1.03 and the ratio of EC to DC = 1.03.
I also found that the angles BPA and EPD have the same angle of
152.44 degrees. Also, the angles BPD and EPA have the same angle
measurement of 27.56 degrees. These angles are alternate interior
angles.**

**This would
make sense since the angles BPA and BPD add up to form 180 degrees.
Does 152.44 + 27.56 = 180? Yes! This is the same case for the
angles EPD and EPA. These two angles add up to form a straight
line having an angle of 180 degrees.**

**Let' s look
at another triangle and make the same comparisons when the point
P is inside the triangle. Click here to explore the location
of P of this triangle as GSP will calculate the values I have
retrieved below for different places of P. Notice that as you
move P around the the inside of triangle ABC, the relationship
between the areas of the triangle ABC and triangle DEF are always
greater than or equal to 4.**

**Notice how
again, for this triangle we get the same relationship where**

**is equal
to 1. If we take 101.18/ 101.18, we get a ratio of 1 between the
two.**

**Below, I
constructed the centroid of the triangle ABC where the three medians
of the triangle ABC intersect. I compared this point to the point
P when the ratio of the area of triangle ABC to triangle DEF is
4. It appears that these points are almost the same point.**

**Now, for
the proof that the ratio of **

** (AF)(BD)(CE)**

**(BF)(CD)(AE)**

**is always
equal to 1.**

**We begin
with the triangle ABC and P any point inside of this triangle.**

**I constructed
a parallal line through the point A to the line BC. Here is what
this lookw like>**

**Next, I see
that the triangles FAH and FBC are similar triangles such that
**

**The triangles
EGA and ECB are also similar triangles such that **

**AG/ BC =
AE / EC = GE / EB. See this illustration below.**

**Triangles
AGP and DBP are also similar triangles where AG/BD = AP/DP = GP/DP**

**See the graph
below for these triangles.**

**Finally,
triangles CDP and HAP are similar where the sides form a ratio
where**

**HA/ DC =
AP/ DP = HP/ CP.**

**Notice how
HA/ DC = AP/ DP and AG/BD = AP/ DP.**

**Therefore,
HA/ DC = AG/ BD and BD/ DC = AG/ AH by cross-multiplying.**

**From above,
we can conclude by saying**

**HA/ BC =
AF/ BF**

**CE/ EA =
BC/ GA**

**BD/ DC =
AG/ AH (what we just found)**

**Multiplying
these gives us**

**(HA/BC)(CE/EA)(BD/DC)
= (AF/BF)(BC/GA)(AG/AH)**

**This simplifies
to 1 and proves that the ratio of **

__(AF)(BD)(CE)__

**(BF)(CD)(AE)**

**is always
equal to 1.**

**This concludes
my investigation of the problem for the final assignment.**