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.
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
(click on the Axiom to see a picture
in Geometers Sketchpad)
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'.
Axiom 2: If AB is congruent to CD and AB is congruent to
EF then CD is congruent to EF.
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'.
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.
Axiom 5: If angle A congruent to
angle B and angle A congruent to angle C, then angle B congruent
to angle C.
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.
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
(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
(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
(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
(b) if and only if
Congruence axiom 1 shows if (2)
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
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
Now, we have exactly one of (1),
(2), or (3) true because exactly one of (a), (b), and (c) is true.
one of the conditions holds:
AB < CD, AB congruent to CD, or AB > CD.
< CD and CD congruent to EF, then AB < EF.
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.
If AB > CD and CD congruent to EF,
then AB > EF.
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.
< CD and CD < EF, then AB < EF (transitivity).
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.
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
First we need another
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.
Exactly one of
the following holds: Angle P < Angle Q, Angle P congruent to
Angle Q, or Angle Q < Angle P
If Angle P <
Angle Q and Angle Q congruent to Angle R, then Angle P < Angle
If Angle P <
Angle Q and Angle Q congruent to Angle R, then Angle P > Angle
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
return to Rives'
EMAT 6680 page