**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.

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

Click here to manipulate a GSP sketch for different locations of P.

You should notice that (AP)(BD)(EC)=(FB)(DC)(EA)

**B**. Now we must
prove why (AP)(BD)(EC)=(FB)(DC)(EA)

First we need to construct parallel lines to produce similar triangles. Below, Line HG is parallel to segment BC

We observe that several triangles appear to be similar. Below is a comparison of similar triangles EGA and ECB.

Similar observations can be made by comparing triangles, FAH and FBC, AGP and DBP, and HAP and CDP. Click here to make the comparisons using a GSP script tool.

(AE/EC)(BF/FA)(CD/DB) = (AG/CB)(CB/AH)(AH/AG) = (AG)(CB)(AH)/(CB)(AH)(AG) = 1

Does this hold true even when P is outside of the triangle? Click here to explore further in GSP.

**C**. Click
here to see a construction showing 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. Notice that the ratio is equal
to 4 when P falls on the centroid of ABC.

This concludes my investigation of the final
assignment. Several of my ideas came from Ceva's Theorem at www.cut-the-knot.org/Generalization/cevas.html.
I would advise you to visit the site for more in depth information.