EMAT 6680 Final Project

Fall 1999

By

Dixie Williford

The problem:

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.

B. Conjecture? Prove it! (you may need draw some parallel lines to produce some similar triangles) Also, it probably helps to consider the ratio (AD)(EB)(CF)/(DB)(CE)(AF).

MY CONJECTURE: For any triangle ABC and for any point P, inside triangle ABC, (AF)( BD)(EC) = (FB)(DC)(EA). Therefore (AD)(EB)(CF)/(DB)(CE)(AF) = 1 for any triangle ABC and any point P inside of triangle ABC. Click here for a GSP sketch in order to explore the construction as well as my conjecture.

PROOF of MY CONJECTURE:

Let's begin by extending segments CD, BF and AE into rays. Also, let's draw in some parallel lines that will be of use.

(Line HA) ll (Ray BF) ll (Line IC)

Now, let's tally up our pairs of similar triangles. (Let "~" mean "is similar to"). Because (Line HA) ll (Ray BF) ll (Line IC) which leads to the AAA similarity postulate, we have that:

Triangle AHD ~ Triangle BPD, Triangle BPE ~ Triangle ICE

Triangle AHC ~ Triangle PFC & Triangle ICA ~ Triangle PFA.

Well, because Triangle AHD ~ Triangle BPD, we know that (AH)/(BP) = (AD)/(BD). Also, because Triangle BPE ~ Triangle ICE, we know that (BE)/(CE) = (BP)/(IC). Well, if X and Y are equivalent and you increase both of them by an equivalent amount, c, then cX =cY. In the same way,

(AD)/(BD) *(BE)/(CE) = (AH)/(BP) *(BP)/(IC) = (AH)/(IC).

Because Triangle AHC ~ Triangle PFC, we know that (PF)/(AH) = (CF)/(AC). Also, because Triangle ICA ~ Triangle PFA, we know that (IC)/(PF) = (AC)/(AF). Therefore,

(PF)/(AH) *(IC)/(PF) = (CF)/(AC)*(AC)/(AF)

Cancelling like terms, we get, (IC)/(AH) = (CF)/(AF).

Remember: (AD)/(BD) *(BE)/(CE) = (AH)/(IC). And we now have that: (IC)/(AH) = (CF)/(AF).

Cancelling like terms we get:

Multiplying both sides by (BD)(CE)(AF), we get:

Can the result be generalized (using lines rather than segments to construct ABC) so that point P can be outside the triangle? Yes. Here is a sketch and measurements showing that the result does indeed generalize.

Triangle ABC was constructed using lines rather than segments. Therefore, even when P lies outside of Triangle ABC, we still get the points of intersection D, E and F. You can see from the measurements that the conjecture still holds.

To satisfy any dounts, click here for a GSP sketch of the above construction. You will be able to see the measurements stay the same even as you manipulate triangle ABC and the point P.

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?

Let the area of Triangle ABC be represented by A(abc) and let the area of Triangle DEF be represented by A(def). I want to show that A(abc)/A(def) is always greater than or equal to 4/1. This means that A(def) is never larger than 1/4 of the area of Triangle ABC.

Let: segment BC = a, segment AC = b, and segment AB = c. Let: segment FE = d, segment DF = e, and segment DE = F. Let a,b,c,d,e and f be non-zero constants.

The area of Triangle ABC = (1/2)ab[sin(angle ACB)].

The area of Triangle DEF = ( 1/2 )ef[sin(angle DFE)].

When is A(abc)/A(def) = 4/1 exactly? I propose that it is exactly when D, E and F are midpoints of AB, BC and AC respectively.

By the Midpoint Connector Therom, we get the following relationships between Triangle ABC and Triangle DEF (these whole only when D, E and F are midpoints of AB, BC and Ac respectively):

2f = 2(DE) = AC, 2(EF) = AB and 2e = 2(DF) = BC

Also, Angle DFE = Angle ACB

Therefore, when D, E and F are midpoints of Triangle ABC, we get that

A(abc) = (1/2)ab[sin(angle ACB)]

A(def) = ( 1/2 )ef[sin(angle DFE)] = (1/2)[(1/2)BC][(1/2)AC][sin(angle DFE)] =

= (1/4)[(1/2)(BC)(AC)sin(angle DFE)].

Because Angle DFE = Angle ACB, we can substitute and get:

A(def) = (1/4)[(1/2)(BC)(AC)sin(angle ACB)] = (1/4)[A(abc)].

Therefore, the ratio of A(abc)/A(def) = A(abc)/(1/4)[A(abc)] = 1/(1/4) = 4. Thus, my proposition is correct. Our ratio equal 4 exactly when D, E and F are midpoints of Triangle ABC.

Is A(abc)/A(def) always greater than or equal to 4?(i.e., is A(def) never larger than 1/4 of A(abc)?) Yes. Click here for a GSP sketch (similar to the one below) and an animation showing that the ratio: A(abc)/A(def) is in fact always greater than or equal to 4.