# Final Project

#### A. Explore (AF)(BD)(EC) and (FB)(DC)(EA) for various triangles and various locations of P.

Below is a picture of the construction. You will also see the measurements and calculations that I want to explore.

Click here to investigate on your own. Change the shape of the triangle and move P around inside triangle ABC. You will see that (AF)(BD)(EC) and (FB)(DC)(EA) will always be the same. Thus,

#### B. Prove it!

I am given the following:

Point P is on the inside of Triangle ABC

I want to prove that :

I will first construct some lines to help me begin my proof. I will extend all sides of the triangle ABC. I will also construct 3 rays with the endpoints at A, B, and C where all pass through P. Then, I will construct a line through point A and parallel to BC.

Consider the two yellow triangles:

I know:

Angle(AEH) = Angle(CEB) because they are vertical angles

Angle(AHE) = Angle(CBE) because they are alternate interior angles

Angle(HAE) = Angle (BCE) because they are alternate interior angles

Therefore, Triangle(AEH) is similar toTriangle(CEB).

By CPCTC, .

Now, consider the two blue triangles below:

I know:

Angle(BFC) = Angle(AFI) because they are vertical angles.

Angle(FBC) = Angle(FAI) because they are alternate interior angles.

Angle(FCB) = Angle(FIA) because they are alternate interior angles.

Therefore, Triangle(FBC) is similar to Triangle(FAI).

By CPCTC, .

Next, consider the two orange triangles below:

I know:

Angle(API) = Angle(DPC) because they are vertical angles.

Angle(PAI) = Angle(PDC) because they are alternate interior angles.

Angle(PIA) = Angle (PCD) because they are alternate interior angles.

Therefore, Triangle(API) is similar to Triangle(DPC).

By CPCTC, .

Finally, consider the two green triangles below:

I know:

Angle(APH) = Angle(DPB) because they are vertical angles.

Angle(PAH) = Angle(PDB) because they are alternate interior angles.

Angle(PHA) = Angle(PBD) because they are alternate interior angles.

Therefore, Triangle(APH) is similar to Triangle(DPB).

By CPCTC, .

So, here is a short recap:

Now, we need to apply some simple algebra.

From the first equation, I can get:

From the second equation, I can get:

From the third and fourth equations, I get:

Now, by substitution,

Cross multiply

Cancel (BC)

And finally...

#### C. Show that when P is inside triangle ABC, the ratio of the areas of triangle DEF is always greater than or equal to 4.

Click here to explore with a GSP sketch of this construction. Notice the ratio of the area of triangle ABC to the area of triangle EFG. It is always greater than or equal to 4.

#### When is it equal to 4?

The ratio of the area of triangle ABC to the area of triangle EFG is equal to 4 iff P is the centroid.