Download Definitions.

MA341
071125
Notes on Chapter 6
This handout gives a summary of topics and definitions in Chapter 6. 1
Definitions.
In the following, we assume that we have a Protractor Geometry (which
means we have a Metric Geometry with Pasch Postulate and a protractor for
all angles.
We are about to add one more axiom and define the Neutral Geometry.
1. Comparisons of segments and angles. Given segments AB and
CD, we say that AB > CD if AB > CD. If AB = CD then we say
AB ∼
= CD.
Given angles ∠ABC, ∠DEF we say that ∠ABC > ∠DEF if m(∠ABC) >
m(∠DEF ) and if m(∠ABC) = m(∠DEF ) then we say ∠ABC ∼
=
∠DEF .
2. Comparison of triangles.
Given a triangle, ABC we shall find it convenient to abbreviate
∠ABC to ∠B and similarly for each of the three vertices.
Given triangles ABC, DEF we say that they are congruent if there
is a bijection F from S = {A, B, C} to T = {D, E, F } such that for
each X ∈ S, we have ∠X ∼
= ∠f (X) and also for X = Y ∈ S, we have
XY ∼
f
(X)f
(Y
).
=
Note that thus the notion of congruence involves checking six corresponding entities to be congruent (the three angles and the three sides).
We shall write ABC ∼
= f (A)f (B)f (C).
Warning.
We shall not use the congruence symbol, unless we have the order of
vertices arranged so that the corresponding entities are congruent. We
will express the congruence by general words “triangles ABC and
DEF are congruent” if we have not specified the correspondence
between the vertices.
Example.
Thus, while it is true that triangles ABC and ACB are always
congruent (when we match all vertices to themselves,) we can say
ABC ∼
= ACB only if AB = AC and m(∠B) = m(∠C), i.e. the
triangle is isoceles with AB = AC. 2
1
Preliminary Version.November 30, 2007
The theorem which says that AB = AC implies ABC ∼
= ACB is known as “pons
asimorum”. It is amusing to look up what this means and why.
This Theorem has a very easy proof due to Pappus, which is valid in a Neutral geometry.
2
1
3. The SAS axiom. In general, congruence of two triangles involves checking six congruences. In the Euclidean Geometry, it can be shown that
a well chosen set of three of the six quantities is enough to deduce the
congruence.
We define:
The SAS Axiom.
Suppose that ABC, DEF are triangles.
If AB ∼
= DE, ∠B ∼
= ∠E and BC ∼
= EF , then ABC ∼
= DEF.
4. Neutral (or Absolute) Geometry.
A Protractor Geometry satisfying SAS axiom is said to be a Neutral
Geometry. Both Euclidean and Poincare Geometries are Neutral. We
shall deduce several congruence theorems and familiar facts assuming
the Neutral Geometry hypothesis.
It is tempting to think that a neutral Geometry shall have all the usual
facts of Euclidean Geometry as true theorems, but that is not the case
unless and until the so-called Parallel Postulate is added in.
• Thus we don’t have the existence of a unique line M through a
point P outside a line L so that L is parallel to M.
• We don’t have the familiar theorem that the sum of the measures
of the three angles of a triangle is a constant 180.
• Many theorems about angles between parallel lines and their transversals are also unavailable.
Theorems in a Neutral Geometry.
1. Pons Asinorum. In a triangle ABC, if AB = AC then m(∠B) =
m(∠C).
Idea of proof. Argue ABC ∼
= ACB using SAS where the corresponding quantities are:
AB = AC, m(∠A) = m(∠A), AC = AB.
It follows that m(∠B) = m(∠C).
2. ASA Theorem. Given triangles ABC, DEF , if
∠A ∼
= ∠D, AB = DE and ∠B ∼
= ∠E
then ABC ∼
= DEF .
Idea of Proof. If AC = DF then we are done by SAS.
Otherwise, assume without loss of generality, that AC < DF .
By segment construction, there is a point G in DF with AC = DG.
By SAS, ABC ∼
= DEG.
2
Then G is seen to be in the interior of ∠DEF and hence ∠DEG <
∠DEF .
But this is a contradiction since
∠DEG ∼
= ∠ABC ∼
= ∠DEF.
Here the first congruence follows from the proven congruence of triangles, while the second one is from hypothesis.
3. ASA implies SAS. It is possible to deduce the SAS statement as a
theorem, if we assume the ASA statement as a hypothesis.
Idea of the proof. Thus, assume we have triangles ABC, DEF
with AB = DE, ∠B ∼
= ∠E, BC = EF .
We wish to prove ABC ∼
= DEF using ASA as a known thoerem.
If ∠C ∼
= ∠F , then we are done by ASA.
We may assume without loss of generality that ∠F < ∠C and construct
−−→
a ray CD such that m(∠BCD) = m(∠F ) and D is in the same side as
←→
A of the line BC.
−−→
From hypothesis, the ray CD meets the crossbar AB, say at some point
G, so that we have A−G−B.
By ASA, we get GBC ∼
= DEF . But then, DE = GB < AB by
betweenness and this contradicts the hypothesis AB = DE.
4. Converse of Pons Asinorum. In a triangle ABC, if m(∠B) =
m(∠C), then AB = AC.
Idea of Proof. As before, argue that ABC ∼
= ACB by comparing
the following ASA elements:
∠B ∼
= ∠C, BC ∼
= CB, ∠C ∼
= ∠B.
The result follows from the congruence of triangles.
5. SSS Theorem.
Given triangles ABC, DEF , if
AB = DE, BC = EF, AC = DF
then ABC ∼
= DEF .
Idea of Proof.
←→
Choose a point Q on the opposite side of A from the line BC such that
QB = DE and ∠QBC ∼
= ∠E.
Then by SAS, deduce that DEF ∼
= QBC.
Then it is enough to prove ABC ∼
= QBC.
←→
The segment AQ must then meet the line BC.
3
Let the intersection point be named G and list the five cases for its
position on the line.
In each case, by repeated application of Pons Asinorum, deduce that
∠A ∼
= ∠Q and then apply SAS axiom using the known equalities:
AB = DE = QB, AC = DF = QC
to finish of the congruence.
6. Relation between SSS and SAS. Unlike the clean equivalence of
ASA and SAS, it is not known if SSS alone is equivalent to SAS.
While SAS implies SSS, the converse can only be proved, for now, with
a few extra assumptions. No counterexamples are known where SSS
holds but SAS does not. Thus, this is an outstanding problem.
7. Perpendiculars. Given a line L and a point P outside it, there is a
line M perpendicular to L containing P .
Recall that this was easy from the protractor axiom when P was on L.
Later we shall show the line M to be unique for a given P and L.
Idea of Proof. Choose two points A, B on L and construct a point
Q on the opposite side of L with respect to P such that m(∠QAB) =
m(∠P AB) and P A = QA.
The segment P Q intersects L at some point M.
First suppose A = M. Then we use SAS to prove that P AM ∼
=
QAM. from this, we deduce that ∠AMP ∼
= ∠AMQ and since they
form a linear pair, each has measure 90 as needed.
In case A = M we get the two congruent angles ∠P AB and ∠QAB to
form a linear pair and hence each is 90 as needed.
This perpendicular will turn out to be unique after the Exterior Angle
Theorem below.
Note that the perpendicular line does meet the line L.
8. Exterior Angle Theorem. Given a triangle ABC choose a point
P such that B−C−P holds.
Then the angle ∠ACP forms a linear pair with ∠C with respect to the
←→
←→
line BC and we call it the linear mate of ∠C along the line BC.
←→
There is a similar linear mate with respect to the line AC. Moreover,
these two mates are congruent, since they form a vertical pair.
The Exterior Angle Thoeorem states that the measure of a linear mate
of ∠C is bigger than the measures of either of the other two angles
∠A, ∠B. These two angles are called the remote interior angles of
the mate.
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Idea of Proof.
3
←→
To prove the result for the mate of ∠C along BC set M to be the
midpoint of AC and choose a point E such that M is the midpoint of
BE.
Then we can argue that AMB ∼
= CME. This gives m(∠A) =
m(∠ECM) and the angle m(∠ECM) is seen to be smaller than our
exterior angle ∠ACP .
9. Uses of the Exterior Angle Theorem. Let L be a line and P a
point outside it. Given points A−B−C on L, we see that for the triangle
P AB the exterior angle ∠P BC at ∠B has bigger measure than that
of the remote interior angle m(∠P AB).
Fix the point B and for any point X = B on L define a function
ψ(X) = m(∠P XB).
−−→
If we let X vary over the interior of the ray CB then we see that
this function ψ(X) steadily decreases as the distance XB increases.
−−→
Moreover, all the values of ψ(X) as X varies over the interior of CB
are less than m(∠P BC).
−−→
In contrast, if we take X to be in the interior of the ray BC, then each
value ψ(X) is less than m(∠P BA). Again, ψ(X) decreases as XB
increases.
It follows that there is a unique B such that P B is perpendicular to L.
We list, without proof, a few more applications.
• SAA Theorem.
This is a generalization of the ASA theorem, where we allow the
two angles to be any two of the three angles, not necessarily the
ones at either end of the side.
In Euclidean Geometry where the sum of all three angle measures
is 180, this trivially equivalent to ASA. In general Neutral geometry, we need a construction.
• Comparison of Sides and Angles.
This is a natural supplement to the result about isoceles triangles.
Consider a triangle ABC. Then
AB > AC iff ∠C > ∠B.
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If unique parallel lines were to exist, then the measure of the exterior mate of ∠C
equals the sum of the measures of the remote interior angles, namely m(∠A) + m(∠B).
However, this is not necessarily true for a Neutral Geometry, hence we have this weaker
statement.
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• Triangle Inequality. Triangle inequality holds in a Neutral Geometry.
Idea of Proof. Given a triangle ABC, we choose D with D−B−C
such that DC = AB +BC and deduce that m(∠D) < m(∠DAC).
It follows that the opposite sides AC and DC satisfy AC < DC =
AB + BC. This proves the triangle inequality.
• Distance from a point to various points of a chosen line.
Let L be a line and P a point outside L. Choose a point B on L
←→
such that the line P B is perpendicular to L.
Let f be a ruler on L such that f (B) = 0. Let P B = d. Define
the function θ from L to such that θ(X) = P X for each point
X on L.
The line is then a union of two rays R+ = {X|f (X) ≥ 0} and
R− = {X|f (X) ≤ 0}. The only common point of the two rays is
B.
It can be shown that θ is an injective function on each of these
rays with range [d, ∞).4 Moreover, for any two points X1 , X2 on
L, we have
|f (X1 )| > |f (X2 )| iff θ(X1 ) > θ(X2 ).
To be continued.
4
Is it surjective?
6