This semester at UGA, I had a course in which we studied Euclidean and non-Euclidean geometry. We spent the first few weeks of class defining undefined terms in Euclids postulates and then discussed the flaws in Euclid's postulates. The course was very interesting, as I had always been taught Euclidean geometry and never thought to question it! We spent the second half of the course looking at what should be added to Euclid's postulates to solidify them. More axioms were needed and we were presented with David Hilbert's system of axioms. (Our text was Euclidean and Non-Euclidean Geometries by Marvin Jay Greenberg) Hilbert's axioms are duvuded into five groups: incidence, betweeness, congruence, continuity, and parallelism. Because class time was limited and many classes were spent discussing just one or two propositions, we focussed mainly on the axioms of incidence, betweeness and congruence. The proofs at many times seemed so trivial they were very difficult to prove. Also, the fact that I have not spent much tme doing proofs and never considered myself very good at them made this course especially difficult. However with practice, I am much more confident in my abilities!

The following are proofs of congruence.

 

SEGMENT ORDERING:

In the following investigation, I will take you through some proofs of congruence by way of segment ordering. Before we get started let's review Hilbert's axioms of congruence:

(click on the Axiom to see a picture in Geometers Sketchpad)

Congruence Axiom 1: If A and B are distinct points and if A' is any point, then for each ray r emanating from A' there is a unique point B' on r such that B' is not equal to A' and AB is congruent A'B'.

Congruence Axiom 2: If AB is congruent to CD and AB is congruent to EF then CD is congruent to EF.

Congruence Axiom 3: If A*B*C, A'*C'*C', AB congruent to A'B', and BC congruent to B'C', then AC is congruent to A'C'.

Congruence Axiom 4: Given any angle BAC (where by definition of "angle," ray AB is not opposite to ray AC), and given any ray A'B' emanating from a point A', then there is a unique ray A'C' on a given side of line A'B' such that angle B'A'C' is congruent to angle BAC.

Congruence Axiom 5: If angle A congruent to angle B and angle A congruent to angle C, then angle B congruent to angle C.

Congruence Axiom 6: If two sides and the included angle of one triangle are congruent respectively to two sids and the incuded angle of another triangle, then the two triangles are congruent.

Throughout the following proofs I will refer back to these axioms.

 

Before we begin this proof we need a definition of <.

Definition: AB < CD (or CD > AB) means there exists a point E between C and D such that AB < CD.

 

Now, let's begin our proofs. I will prove a series of four propostions. All of which seem very obvious to us who have studied Euclidean geometry. These proofs incorporate some of Hilbert's congruence axioms, betweeness axioms, and propositions (which I will define as we go). The proof of segment ordering has four parts. In part A, we must first prove that when dealing with segments, one must be greater than the other, less than the other, or congruent to the other. Once we show this the other proofs follow.

Part A

Exactly one of the following holds (trichotomy):
AB<CD, AB congruent to CD, or AB > CD.

We will refer to the above cases now by (a), (b), or (c) and the three cases above by
(1) AB < CD


(2) AB congruent to CD


(3) AB > CD

 

 

By congruence axiom 1, there is a unique point B' on CD so that AB ? CB'.
We know that B' ? C , so by definition of ray at least one of the following is
true:
(a) C * B' * D

(b) B' = D

(c) C * D * B'

 

Now, we would like to see that exactly one of (a), (b), or (c) is true:

(a) and (c) cannot both be true, by betweeness axiom 3
(a) and (c) cannot both be true by betweeness axiom 1
(b) and (c) cannot both be true by betweeness axiom 1.
Therefore, exactly one of (a), (b), and (c) are true.

Next, we want to see that
(a) if and only if (1)
(b) if and only if (2)
(c) if and only if (3)

Let's look now at each of these cases and see if each condition holds.

(a) if and only (1):

This is precisely the definition of <.

(b) if and only if (2):

Congruence axiom 1 shows if (2) then (b)
To see if (b), then (2), assume D = B'. Then CB' congruent to CD, by congruence axiom 2 and AB congruent to CB'. So, AB congruent to CB' by congruence axiom 2.

(c) if and only (3):

To see if (c) , then (3), assume C * D * B'. Then, by proposition 3.12 (given
AC congruent to DF, for any point B between A and C there is a unique point E between D and F, such that AB congruent DE, there is a unique point D' between A and B so that CD congruent to AD'. So, by definition of <, CD < AB.
Now, to see if (3), then (c), assume CD < AB. Then there exists D' between A and B so that AD' congruent to CD. We know AB congruent to CB', so also by proposition 3.12, there is a point D'' so that C*D''*B and AD' congruent to CD''. Then by congruence axiom 2, CD congruent to CD''. So by congruence axiom 1, D = D'' and therefore C*D*B'.

Now, we have exactly one of (1), (2), or (3) true because exactly one of (a), (b), and (c) is true.

Therefore, exactly one of the conditions holds:
AB < CD, AB congruent to CD, or AB > CD.

 

 

Part B

If AB < CD and CD congruent to EF, then AB < EF.

see sketch

Proof:
Given AB < CD, they by definition of <, there exists a point G on CD such that C*G*D and AB @ CG. By proposition 3.12, there exists a point G' on EF, such that CG congruent to EG'. Then by congruence axiom 2, AB congruent to EG'.
Therefore, AB < EF by definition.

Part C

If AB > CD and CD congruent to EF, then AB > EF.

see sketch

Proof:
CD < AB. Then by definition, there exists a point G on AB such that A*G*B and AG congruent to CD. Since CD congruent to EF and CD congruent to AG then by congruence axiom 2,
EF congruent to AG.

Therefore, by definition, AB > EF.

 

Part D

If AB < CD and CD < EF, then AB < EF (transitivity).

see sketch

Proof:

There is a point such that C*G*D and AB congruent to CG by definition of <. Also by definition of <, there is a point D' between E and F such that CD congruent to ED'. There is a unique point M between E and D' so that CG congruent to EM, by proposition 3.12. Now by congruence axiom 2, AB congruent to CG congruent to EM. Now, since E*D'*F and E*M*D', then E*M*F and by definition of <, AB < EF.

The END!

 

There is a similar proposition dealing with angles. Take some time and see if you can use the proof above to guide you. The proofs are very similar, just change a few axioms and a bit of notation and you'll have it.

Ordering of Angles:

First we need another defintion:

Definition: Angle ABC < Angle DEF means there is a ray EG between ray BA and ray BC, ray EH between ray ED and ray EF, angle CBG congruent to Angle FEH and Angle ABC congruent to Angle DEF. Then Angle GBA is congruent to Angle HED.

part A

Exactly one of the following holds: Angle P < Angle Q, Angle P congruent to Angle Q, or Angle Q < Angle P

part B

If Angle P < Angle Q and Angle Q congruent to Angle R, then Angle P < Angle R.

part C

If Angle P < Angle Q and Angle Q congruent to Angle R, then Angle P > Angle R.

part D

If Angle P < Angle Q and Angle Q < Angle R, then Angle P < Angle R.

 

Link to geometers sketchpad so you can draw these propositions

 

 

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