**Problem**: 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.

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

b.) Conjecture? Prove it! Can the result generalize(using lines rather than segments to construct ABC) so that point P can be outside the triangle?

c.) Show that the ratio of the areas of ABC and DEF is always less than or equal to 4. When is it equal to 4?

Below is the construction of triangle ABC with a point P selected in the inside of the triangle. Then I connected AP, BP, and CP to their intersections with the opposite sides at points D, E, and F.

Once I made this construction I explored various triangles and various locations for P. Through this exploration I realized that (AF)(BD)(EC) and (FB)(DC)(EA) are always equal. I took this a step further to say that the ratio of (AF)(BD)(EC) to (FB)(DC)(EA) is always 1. I explored this for different locations of P in different triangles and this never changed. The ratio was always 1. I took this a step further and constructed the same above triangle but this time with lines instead of segments. By doing this I saw that I could take my above statement even further. I found that even when I moved my point P outside the triangle the ratio of (AF)(BD)(EC) to (FB)(DC)(EA) is still 1.

My conjecture for part B is that the ratio of (AF)(BD)(CE)/(BF)(CD)(AE) = 1.

I will begin doing this by using similar triangles. My first step was to construct similar triangles by constructing parallel lines to line segment AD. These parallel lines will go through the vertices of B and C. This construction can be seen below.

From the above construction, you can see that my parallel lines are the dotted green lines. The First set of triangles that we will look at are triangle APF and triangle BFH. We will show that they are similar triangles. For two triangle to be similar they must have congruent corresponding angles and corresponding sides must be in proportion. In these two triangles, angle A is congruent to angle B and angle H is congruent to angle P because the are alternate interior angles. Angle F is congruent to angle F because they are vertical angles. The sides that are in proportion are the following.

(AP/BH) = (FA/BF)

The second set of triangles we want to prove are similar is triangle AEP and triangle CTE. By following what I did above, angle T and angle P are congruent and angle C and angle A are congruent because both sets are alternate interior angles. Also, angle E is congruent to angle E because they are vertical angles. The ratios that are in proportion are the following.

(EC/EA) = (CT/AP)

A third set of triangles that we will prove are similar are triangles HBC and triangle PDC. Again angle D is congruent to angle B and angle P is congruent to angle H because they are alternate interior angles. Also, angle C is congruent to angle C because they are vertical angles. The sides that are in proportion are the following.

(BH/DP) = (BC/BD)

Finally , the last set of triangles that we will show are similar are triangle TBC and triangle PDB. Angle D and angle C are congruent and at the same time angle P is congruent to angle T because they are alternate interior angles. Also, angle B is congruent to angle B because they are vertical angles. The sides that are in proportion are the following.

(TC/PD)= (BC/DB)

So we can see these better let me list them below. As you can see we have four sets of ratios to work with. They are as follows.

1. (FA/BF) = (AP/BH)

2. (EC/EA) = (CT/AP)

3. (BH/DP) = (BC/BD)

4. (TC/PD)= (BC/DB)

Before we multiply these together to prove my conjecture there is one step we need to do. We can see that 3 and 4 are set equal to the same thing which allows us to set the following equal to each other.

(BH/DP) = (TC/PD)

From this we can rearrange the above proportion as (PD/PD) = (BH/TC).

Now we are ready to multiply our ratios together.

(FA/BF) * (EC/EA) * (PD/PD) = (AP/BH) * (CT/AP) * (BH/TC) = 1.

Therefore this can be generalized to say that if three lines intersect at one point the ratio [(AF)(BD)(CE)]/[(BF)(CD)(AE)] =1.

Click here to explore the ratio of the areas of triangle ABC and DEF.

From the above GSP exploration you can see that the ratio of the area of triangle ABC to triangle DEF is always greater than 4. There was only one time when I got that ratio to equal four. This was when triangle DEF was the medial triangle. We know from previous assignments that the medial triangle is similar to the original triangle and that its area is (1/4) the area of the original triangle. Therefore, if you flip the ratio then you can say that the area of triangle ABC is 4 times the area of the medial triangle.For all other cases of where P is inside the triangle the ratio of the area of triangle ABC will be greater than 4.