Final Assignment

Is Concurrency a Coincide ?

by

A. Kursat ERBAS

Almost everybody has been thought (or has studied) the Euclidean Goemetry in high school or college years. In any triangle triangle, altitudes, medians, angle bisectors and perpendicular bisectors are concurrent. In fact, concurrency of lines or segments are very fundemental part of the Euclidean Geometry in general. In this study, we will try to investigate conccurrency of segments passing through three vertices of a triangle. This investigation may help us to answer the question:

What are the required and sufficient conditions for a concurrency of three lines passing through three vertices of a triangle?

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 as in Figure 1.

 Figure 1

Using GSP, it is not difficult to observe the following fact for the above constrcution (See Figure 2).

 Figure 2

The observation shows that in every triangle ABC, (AF).(BD).(EC) = (FB).(DC).(EA). In other words, their raio is "1".

Click Here to observe this fact with a GSP demonstration showing it for various triangles and points in the triangles

Of course we should prove this observation for a universal convenience.

Conjecture: Let ABC be a triangle and P be a point inside it. Let the lines AP, BP, and CP intersects the sides BC, AC, and AB in points D, E, and F respectively. Then (AF).(BD).(EC) = (FB).(DC).(EA)

Proof: Let k be the line passing through A and parallel to BC. Let H be the intersection point of the line k and the line passing through C and P. And, let G be the intersection point of the line k and the line passing through B and P (See Figure 3).

 Figure 3

By Angle-Angle Similarity Theorem;

the triangles AFH and BFC are similar since m(FHA)=BCF) and m(HAF)=m(CFB).

Similarly by Angle-Angle Similarity; AEG~CEB, APG~DPB, APH~DPC.

From AFH~BFC, and AEG~CEB: AF/BF = AH/BC and AE/CE = AG/CB.

Also from APG~DPB; AG/DB = AP/DP and from APH~DPC, AH/DC = AP/DP. Therefore AG/BD = AH/DC which gives us BD/DC = AG/AH.

Then multiplying the proportions of the sides of the triangle ABC we may get:

(AF/FB)(BD/DC)(CF/EA) = (AH/BC)(AG/AH)(BC/AG) = 1

In other words, (AF).(BD).(EC) = (FB).(DC).(EA) holds if given triangle ABC, three lines AD, BE, CF intersect at a single point P.

To prove the conjecture in other directions (i.e. drawing parallel lines from B to AC and from C to AB) is almost as same as the above.

One major question may come to one's mind that

"Can the result be generalized (using lines rather than segments to construct ABC) so that point P can be outside the triangle?" (See Figure 4)

 Figure 4

A GSP construction for the above question shows that the result can be generalize in such a way:

Theorem: Let ABC be a trinagle and P be a point (inside,outside or on it). Let the lines AP, BP, and CP intersects the lnes BC, AC, and AB in points D, E, and F respectively. Then

(AF).(BD).(EC) = (FB).(DC).(EA)
Click Here for a GSP sketch for the above theorem

Onether interesting observation is about the relation between the areas of six triangles: PCE, PBD, AFP, APE, PDC, FPE. The observation is

 Figure 5

The nicest thing here is that the property holds whreever P is.

Click Here to observe this with GSP (if P is inside ABC)

Click Here to observe this with GSP (general form)

By GSP, another interesting observation may be 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 (See Figure 6a).

 Figure 6a Figure 6b

Click Here for a GSP file for this observation

When is it equal to 4?

A simple GSP observation may lead that when P is the centroid (G) of the triangle ABC, the ratio is equal to 4.

The proof in general is:

Statement: 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.

Then, the ratio of the areas of triangle ABC and triangle DEF is always greater than or equal to 4.

Proof: Let a = BD/BC, b = CE/AC, c = AF/AB. Then Area(DEF)/Area(ABC) = 2abc. So we want to show that abc is less than or equal to 1/8.

Let x = a/(1-a), y = b/(1-b), z = c/(1-c). The Conjecture we found above says that xyz = 1.

Now a = x/(1+x), b = y/(1+y), c = z/(1+z). So we want to show that xyz/(1-x)(1-y)(1-z) is less or equal to 1/8. In other words we want to show that

(*) (1-x)(1-y)(1-z) is greater than or equal to 8, provided that x, y, and z are positive and xyz = 1.

Using calculus, it's easy to show that if x is positive then x + 1/x is greater than or equal to 2. Using this fact and a little algebra, we get (*).

Note: Thanks to Dr. Macrory (clint@math.uga.edu) for his help on this proof.

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This page created December 6, 1999

This page last modified December 21, 1999