A. In triangle ABC, we select any point P inside the triangle and draw the lines AP, BP, and CP extended to their intersections with the opposite sides D, E and F respectively.

We eplore (AF)(BD)(EC) and (FB)(DC)(EA) for various triangles and various locations of P.

In the above triangle:

These products seemed to remain the same for all triangles no matter where P is located.

B. Conjecture: (AF)(BD)(EC) = (FB)(DC)(EA)

C. We now look at the ratio of the area of triangle ABC to the area of triangle DEF.

We can click **here** to see
that the ratio of ABC to DEF is always greater or equal to 4.

This can be explained by the greatest area of DEF is when DEF is the medial triangle. When this occurs, ABC is cut into 4 congruent triangles, thus having a ratio of 4 to 1.