Consider the triangle ABC below where point P is randomly selected in the interior of triangle ABC. If I draw lines from each of the vertices through P, I get the following construction.

Points D,E, and F are where the lines AP, BP, and CP intersect the opposite
side of the triangle, respectively. Once I had my construction, I wanted
to explore the relationship of (AF)(BD)(CE) and (FB)(DC)(EA). In other
words, I wanted to explore the relationship of the product of these segment
lengths for various triangles and various P's.

By using GSP and exploring different types of triangles and different points
for P, I found that (AF)(BD)(CE)=(FB)(DC)(EA). Thus, making the ratio one
to one.

**Proof of the conjecture
**To prove this conjecture, I constructed parallel lines on the previous
triangle to form similar triangles.

I found the following similar triangles:

triangle BPD is similar to triangle BRC

triangle CPD is similar to triangle CNB

triangle APE is similar to triangle CRE

triangle AFP is similar to triangle BFN

triangle CPD is similar to triangle BDK

One can prove the above conjecture by using the above similarities and algebra.

The following equalities were found from the above similarities.

BD/BC=BP/BR=PD/RC

CP/CN=CD/CB=PD/NB

RE/PE=EC/AE=RC/AP

AF/FB=AP/NB=PF/FN

DK/PD=BD/CD=BK/CP

From these equalities, I concluded the following:

AF/FB=AP/NB implies that (AF)(NB)=(FB)(AP)

(NB)=(PD)(CB)/(CD)

(AP)=(AE)(RC)/(EC)

(AF)(PD)(CB)/(CD)=(FB)(AE)(RC)/(EC)

(AF)(PD)(CB)(EC)=(FB)(AE)(RC)(CD)

From the first similarity, we know that (PD)=(BD)(RC)/(BC).

When I subsituted this in for (PD) I got the following:

(AF)(BD)(RC)(CB)(EC)/(BC)=(FB)(AE)(RC)(CD)

The (BC) will cancel and the (RC) will cancel. Thus

(AF)(BD)(EC)=(FB)(AE)(CD).

Thus, concluding the proof.

Now, let's see if this result can be generalized so that point P can be
outside of the triangle ABC.

To investigate this, we need to use lines instead of segments to construct
our triangle ABC. I investigated this using the following triangle. I
used the circle to animate P along the outside of triangle ABC.

As I animated P along this circle, I noticed that the ratio (AF)(BD)(EC)/(BF)(DC)(EA)
is still equaled to one.

Using the first triangle, I am going to investigate the ratio bewteen triangle
ABC and triangle DEF. I am using the following construction to perform
my investigations.

I'm going to investigate the ratio of the area between these two triangles. During my investigations, I found something that was very interesting. I noticed that the ratio of the area was either 4 or greater. The ratio is equaled to 4 when the areas of each of the four interior triangles are the same. Observe the following diagram.

The areas are equal when P is the centroid of triangle ABC. Thus, when
P is the centroid the ratio of the areas of triangles ABC and DEF is 4.
When P is anywhere beside the centroid, the ratio is greater than 4.

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