Download Introduction to Neutral Geometry

Introduction to Neutral Geometry
( Sections 3.1, and 3.2 )
Neutral Geometry is the axiom system with the same undefined terms as those of Euclidean Geometry
but using the SMSG Postulates 1 – 15 as the only axioms. One can think of Neutral Geometry as the
geometry which results from using as the axioms only Euclid’s first four postulates (and those statements
he assumed as true without proof in his book “Elements” without listing them as postulates). In
particular, Neutral Geometry has no parallel postulate such as Euclid’s 5th Postulate or such as the
SMSG Postulate 16 (Playfair’s Postulate).
Since the SMSG Postulates 1 – 15 are axioms of Hyperbolic Geometry (as well as axioms of Euclidean
Geometry), the proofs of theorems in Neutral Geometry also prove that those statements are true in
Hyperbolic Geometry and they prove that those statements are true in Euclidean Geometry.
Similarly, these theorems proven using only Euclid’s first four postulates are true in both Euclidean
Geometry and Hyperbolic Geometry.
Among these first 15 SMSG Postulates are Postulate 4 (The Ruler Placement Postulate) and Postulate
12 (The Angle Construction Postulate). In proofs, logical arguments based on Postulate 4 alone allow
one to conclude the existence and uniqueness (by SMSG Postulate 4) of points that are any specified
distance away from any specified point along any specified line in any specified direction along the line.
One can also conclude that, given two distinct points A and B, there exists (by SMSG Postulate 4) a
unique distinct third point C such that the ratio (AC) / (CB) is equal to any specified positive real
number. In proofs in this class, assertions of the existence or uniqueness of points such as these may
have the phrase “by SMSG Postulate 4” or the phrase “by the Ruler Placement Postulate” as their
justifying reason.
Also, sections of proof based on SMSG Postulate 12 alone allow one to conclude the existence and
uniqueness (by SMSG Postulate 12) of angles of any specified measure with any specified ray as one side
of the angle with the other ray in a specified one of the two half-planes determined by the line containing
the specified ray. In proofs in this class, assertions of the existence or uniqueness of angles such as these
may have the phrase “by SMSG Postulate 12” or the phrase “by the Angle Construction Postulate” as
their justifying reason.
These practices are illustrated in the proof of the following theorem:
Theorem 3.2.2: (i) Every line segment has exactly one midpoint;
(ii) Every angle has exactly one bisector.
Proof: (i) Let AB be any line segment. By SMSG Postulate 4, there exists a unique point C on line

AB and between A and B such that AC = ½ AB . Since C is between A and B,
CB = AB – AC = AB – ( ½ AB ) = ½ AB
Therefore, AC = CB and  C is the unique midpoint of AB .

(ii) Let  ABC be any angle. By SMSG Postulate 12, there exists a unique ray BP in the interior

of  ABC (and so, also, in the half-plane determined by line AB that contains C) such that m(  ABP )
= ½ m(  ABC ) . By SMSG Postulate 13, the Angle Addition Postulate, since P is in the interior of
 ABC ,
m(  PBC )
=
m(  ABC ) – m(  ABP )
=
m(  ABC ) – ½ m(  ABC )

=
½ m(  ABC ) .
Therefore, m(  ABP ) = m(  PBC ) and  BP is the unique angle bisector of  ABC . QED
2
Definition: Two angles are supplementary if the sum of their measures is 180°. When two angles are
supplementary, each one is called a supplement of the other.
Two angles are complementary if the sum of their measures is 90°. When two angles are
complementary, each one is called a complement of the other.
Theorem 3.2.3 (The "Common Supplements and Complements" Theorem):
Supplements and complements of the same or congruent angles are congruent.
Theorem 3.2.3 (Restated for clarity)
( 1 ) Supplements of congruent angles are congruent.
( 2 ) Complements of congruent angles are congruent.
( 3 ) Supplements of the same angle are congruent.
( 4 ) Complements of the same angle are congruent.
Proof of (1): Let  ABC ,  DEF ,  PQR and  XYZ be any four angles such that  ABC is a
supplement of  PQR and  DEF is a supplement of  XYZ and  PQR   XYZ .
[ We need to prove that  ABC   DEF . ]
By definition of “supplement,”
m   ABC  + m   PQR  = 180° ,
so m   ABC  = 180° – m   PQR  , and
m   DEF  + m   XYZ  = 180° ,
so m   DEF  = 180° – m   XYZ  .
By definition of “congruent,” m   PQR  = m   XYZ  .
 By substitution, m   ABC  = 180° – m   XYZ  .
  ABC   DEF .
 m   ABC  = m   DEF  .
 Supplements of congruent angles are congruent.
[ QED for (1) ]
Proof of (2): Exercise.
Proof of (3): Let  ABC ,  DEF , and  PQR be any three angles such that  ABC is a
supplement of  PQR and  DEF is also a supplement of  PQR .
[ We need to prove that  ABC   DEF . ]
By definition of “supplement,”
m   ABC  + m   PQR  = 180° ,
so m   ABC  = 180° – m   PQR  , and
m   DEF  + m   PQR  = 180° ,
so m   DEF  = 180° – m   PQR  .
 By substitution,  m   ABC  = m   DEF  , since both are equal to 180° – m   PQR  .
  ABC   DEF .
Proof of (4): Exercise.
 Supplements of the same angle are congruent.
QED
[ QED for (3) ]
3
Theorem 3.2.4 (The "Vertical Angles" Theorem): Vertical angles are congruent.
Proof: Exercise.
Theorem 3.2.5 (Pasch’s Axiom): If a line l intersects  PQR at a point S such that P-S-Q (and PQ
does not lie along line l ) , then line l intersects PR
or RQ .
Proof: Suppose line l is a line that intersects  PQR at a point S such that P-S-Q (and PQ does not
lie along line l ) . [ We need to show that line l intersects PR or RQ . ]
[To prove that line l intersects PR or RQ , we assume that
R
line l does not intersect PR and prove that l intersects RQ . ]
Assume that line l does not intersect PR .
P
l
S
Q
[We prove that line l intersects RQ with a proof-by-contradiction.]
Suppose, by way of contradiction, that line l does not intersect RQ .
Since line l does not intersect PR , P and R are
on the same half-plane side of line l .
R
Since line l does not intersect RQ , R and Q are
on the same half-plane side of line l .
P
l
S
Q
Therefore, P and Q are on the same side of line l .
 PQ is contained in one half-plane side of l .
Therefore, PQ does not intersect line l , which contradicts the fact that a line l intersects PQ point S
with P-S-Q .
Therefore, we can conclude that line l intersects RQ .
Therefore, line l intersects PR or RQ .
QED
4
Theorem 3.2.6 (The "Crossbar" Theorem): If X is a point in the interior of  UVW , then UX
intersects WV at a point Y such that W-Y-V .
W
Proof: Exercise.
Y
X
V
U
Theorem 3.2.7 (The "Isosceles Triangle" Theorem OR The "Two Congruent Sides" Theorem):
If two sides of a triangle are congruent, then the angles opposite those sides are congruent.
Q
Proof: Suppose that  PQR is any triangle such that QP  QR (so
that  PQR is an arbitrary isosceles triangle) . [ We need to show
that  QPR   QRP . ]
By Theorem 3.2.2,  PQR has an angle bisector, say QX for
some point X in the interior of  PQR (which can be assumed
to be in the interior of  PQR ).
X
P
T
R
By the Crossbar Theorem, Theorem 3.2.6 applied to  PQR , ray QX
intersects PR at some point T with P-T-R.
  PQT   RQT by definition of “angle bisector”.
By the initial supposition, PQ  RQ .
Also, QT  QT since every segment is congruent to itself.
Also, the included angles are congruent:  PQT   RQT .
Therefore, by SMSG Postulate 15 (The SAS Congruence Condition),  PQT   RQT .
Therefore,  QPT   QRT because Corresponding Parts of Congruent Figures are congruent.
That is to say,  QPT   QRT by C P C F .
Since  QPR   QPT and  QRP   QRT ,
 QPR   QRP by substitution.
Q E D
[ Recall that you must explicitly establish the congruence of corresponding sides and the congruence of
the corresponding included angles before you can apply the Side-Angle-Side Congruence Condition.
There are similar requirements for the application of the other Congruence Conditions.]
5
Theorem (The "Congruent Supplementary Angles" Theorem):
If two angles are both supplementary angles and congruent angles, then they are right angles.
Proof: Exercise. [This is a direct result of the fact that if X + X = 180, then X = 90.]
Definition: A Conditional Statement is an IF-THEN statement of the form: “IF P, THEN Q” .
The CONVERSE of this conditional statement is the reversed statement: “IF Q, THEN P”
The statement “P IF AND ONLY IF Q” means
“ ‘IF P, THEN Q’ is true
AND
‘IF Q, THEN P’ is true” .
The proof of the statement “P IF AND ONLY IF Q” is always in two parts:
One part presents the proof of the statement “IF P, THEN Q”,
and the other part presents the proof of “IF Q, THEN P”
(or vice versa).
Definition: Given a segment AB , the perpendicular bisector of segment AB is the line which is
perpendicular to line AB and which passes through the midpoint of AB .
Theorem 3.2.8 (The "Perpendicular Bisector" Theorem):
A point is on the perpendicular bisector of a line segment if and only if it is equidistant from the
endpoints of the line segment.
Proof:
[Part I: We prove first that, if a point is equidistant from the endpoints of a line segment, then that
point is on the perpendicular bisector of the line segment.]
Let AB be a line segment and suppose that point P is equidistant from A and B. Thus, PA = PB .
[ We need to show that P is on the perpendicular bisector of AB . ]
Either P is on line AB or P is not on line AB .
Case 1: Assume that P is on line AB . Then, since P is equidistant from A and B, P is the midpoint
of segment AB . P is on the perpendicular bisector of AB since the midpoint of a segment is, by
definition, on the perpendicular bisector of that segment.
P is on the perpendicular bisector of AB in Case 1.
6
Case 2: Assume that P is not on line AB and recall that PA  PB .
Let T be the midpoint of AB and construct line segment PT
(Note: Once points P and T have been defined, the statement “construct the line segment PT ” is not
really necessary because the segment PT exists already. “Construct the line segment PT ” is said
mainly to tell the reader of the proof to draw the segment in the diagram if it hasn’t been drawn
already.)
Since PA  PB ,  PAB   PBA by Theorem 3.2.7 (The Isosceles
Triangle Theorem).
P
Thus,  PAT   PBT by substitution.
Since T is the midpoint of AB , AT  BT .
Thus,  PTA   PTB by SAS, and so,  PTA   PTB by C P C F .
Now,  PTA and  PTB are congruent and, by Postulate 14,
supplementary.
A
B
T
By the (The Congruent Supplementary Angles Theorem) above,
they are both right angles. Thus, PT is perpendicular to AB . PT also passes through the midpoint
T of AB . Therefore, PT is the perpendicular bisector of AB .
Thus, since P is on the line PT , P is on the perpendicular bisector of AB .
P is on the perpendicular bisector of AB in Case 2.
P is on the perpendicular bisector of AB in general since P is on the perpendicular bisector of AB in
all cases.
End of Part I of the proof.
Part II of the proof is left as an exercise.
Q E D.
Definition 3.2.4: The angle formed by extending one side of a triangle (thus producing an angle that is
both supplementary to and adjacent to an interior angle
of the triangle) is called an
exterior angle of the triangle.
For example, in the triangle to the right, side AC is extended to
point P . The resulting angle  PAB is an exterior angle
of the triangle and it is supplementary to and adjacent to
the interior angle  BAC .
B
P
A
C
7
Theorem 3.2.9, The Exterior Angle Theorem:
Each exterior angle of a triangle is greater in measure than either of the non-adjacent
interior angles of the triangle.
A
F
E
B
C
Proof:
Suppose
D
ABC is a triangle.
We can extend BC along ray BC to point D. Locate point E as the midpoint of AC so
AE = CE.
We can extend BE along ray BE to a point F such that BE = EF and so
E is the midpoint of BF .
Since vertical angles are congruent, AEB  CEF .
Now, AE  CE , BE  FE , and AEB  CEF , so
BEA 
 BAE  FCE by CPCF.
Since m(DCF) > 0 and m(DCA) = m(DCF) + m(FCA) ,
m(DCA) > m(FCA).
But,
mFCE = mBAE since BAE  FCE.
Therefore, m(DCA) > m(BAC) = m(BAE).
A similar argument proves that m(DCA) > m(ABC) .
QED
FEC by SAS.