```
Final Write-Up
Triangles, Triangles, Triangles!
By Kevin Mylod
```
```Our final assignment consists of three parts.  Below is the initial investigation:

Part 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.```

Below is the construction of DABC with arbitrary point, P, inside. From P came points D, E, and F. Go ahead and click on the triangle below to explore (AF)(BD)(EC) and (FB)(DC)(EA) for various triangles and various locations of P (inside the triangle).

You see, from the GSP construction, that the above relationship, that is (AF)(BD)(EC) = (FB)(DC)(EA), is true no matter where you move A, B, C, and in the plane. Even moving P inside the triangle yields a true statement. In the next part, we will see that (AF)(BD)(EC) = (FB)(DC)(EA) holds true for all locations of P, inside of DABC or out. We will also prove why this relationship is true for all triangles ABC and locations of P.

```

Part B:
I.  Conjecture? Prove it! (you may need draw some parallel lines to produce some similar
triangles) Also, it probably helps to consider the ratio```

This ratio will help us prove why (AF)(BD)(EC) = (FB)(DC)(EA) because, by dividing both sides by (FB)(DC)(EA), we get . We can show this ratio through several similar triangles created by constructing lines through A and B (shown by the blue dashed lines) parallel to line CP. Point X results from the intersection of lines AP and AX and point Y results from the intersection of lines BP and BY.

To see the similar triangles created from the picture above, we've isolated them in the figures below. The similar triangles are featured in green.

```Figure 1                        Figure 2                                    Figure 3                       Figure 4
```
`                 DAFP ~ DABY                    DBFP ~ DBAX                    DBDY ~ DCDP                  DCEP ~ DAEX`

Proof:

The definition of similar triangles tells us that each side of one triangle is proportional in length to the corresponding of the other triangle. Thus, the following ratios come out of this definition:

```Figure 1                 Figure 2                 Figure 3                 Figure 4
```

If we cross multiply the first two ratios we get (AF)(BY) = (BF)(AX) = (AB)(FP). From these two products come the following ratio:

If we take this ratio along with the ratios from figures 3 and 4, and multiply them together we get the following:

Notice that everything on the right side of the equation will cancel out, leaving the left side of the equation equal to one (1). Thus, our proof shows that the product (AF)(BD)(EC) = (FB)(DC)(EA) is true for all triangles, ABC, and for all locations of arbitrary point, P.

```                         II.  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.```

The answer is yes, the result can be generalized. The above work is under the assumption that DABC is constructed out of segments. The constuction of DABC with lines will allow us to move point P anywhere inside or outside of the triangle without the loss of any measurements or products.

```

Part 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?

Click on the above phrases in bold to see the GSP scetches for the answers.

Well, this is almost the end of the our journey.  It's now time to WRAP THINGS UP!