More on Neutral Geometry
I
(Including Section 3.3)
( "NIB" means
"NOT IN BOOK" )
Theorem (NIB),
The "The Adjacent Supplementary Angles" Theorem
(Converse of Postulate 14) :
If two adjacent angles are supplementary, then they form a linear pair.
Proof:
Suppose that
ABD
∠
and
DBC
∠
are two adjacent angles (with common
ray
BD
----→
)
are supplementary.
(We need to show that they form a linear pair.)
By Postulate 4 (The Ruler Placement Postulate),
there is a point on line
AB
←---→
such that A – B – F .
Thus, since F and C are both in the half-plane side of
BD
←---→
opposite the half-plane containing A,
F and C are on the same side of
BD
←---→
.
Now,
ABD
∠
and
DBF
∠
form a linear pair by definition of “linear pair”.
By Postulate 14,
ABD
∠
and
DBF
∠
are supplementary.
By the supposition above,
ABD
∠
and
DBC
∠
are also supplementary.
Therefore,
DBC
DBF
∠
2245 ∠
./
∴
(
)
(
)
m
DBC
m
DBF
∠
=
∠
.
We see then that ray
BC
----→
and
BF
----→
are both rays in the same half-plane side of
line
BD
←---→
such that each ray combines with ray
BD
----→
to produce angle with the
same measure
r
=
180
°
–
(
)
m
ABD
∠
.
By Postulate 12 (The Angle Construction
Postulate), there is one and only one ray in that same half-plane side of
BD
←---→
which
unions with
BD
----→
to produce the angle with that particular angle measure, r
=
(
)
(
)
m
DBC
m
DBF
∠
=
∠
.
By the uniqueness of this ray, we can conclude that
BC
----→
=
BF
----→
and so,
DBC
DBF
∠
=
∠
.
Therefore,
∠ g1827g1828g1830
g1853g1866g1856
∠ g1830g1828g1829
form a linear pair.
Q E D
B
A
D
C
F

2
Section 3.3:
Triangle Congruence Conditions
Theorem 3.3.1 (Angle-Side-Angle Congruence Condition):
If, in two triangles, the vertices of one triangle can be put into one-to-one correspondence with
the vertices of the other triangle such that:
Two angles and the included side of one triangle are congruent to the corresponding angles
and included side of the other triangle,
Then, the correspondence of triangle vertices is a congruence and the two triangles are
congruent.
Proof:
Let
ABC
∆
and
DEF
∆
be two triangles and consider the correspondence of vertices
ABC
barb2leftbarb2right
DEF .
Suppose that
CAB
FDE
∠
2245 ∠
,
and
CBA
FED
∠
2245 ∠
,
and
AB
DE
2245
.
[We need to show
ABC
DEF
∆
2245 ∆
. ]
[ We do this by showing that
AC
DF
2245
, and then we apply Postulate 15 (SAS) to conclude that
ABC
DEF
∆
2245 ∆
.]
Suppose, by way of contradiction, that
AC
and
DF
are not congruent .
∴
AC
≠
DF .