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Proofs o
f Some EPP Equivalences (in Neutral Geometry)
Recall the correct wording of the EPP (Euclidean Parallel Postulate):
For every line
l
and every point P not on line
,
there is one and only one line
m
that contains P and is
parallel to
l
.
Note that this wording reminds us that, in proving that the EPP is true under certain assumed conditions, we
first define an arbitrary line
l
and an arbitrary point P not on
l
.
Then, we have two tasks to complete:
1) We must prove that there exists at least one line
m
that contains P and is parallel to
l
.
[This is accomplished by applying the (NIB) "Common Perpendicular" Theorem and then the "Common
Perpendicular makes Parallel Lines" Theorem (Corollary 3.4.2) .]
2) We must prove that there does not exist a second line
n
that contains P and is parallel to
l
.
[This is accomplished by using proofbycontradiction: First, assume that there exists a second line
n
,
n
m
, that contains P and is parallel to
l
; then, use logical argument to prove that a contradictory situation
exists under that assumption.]
This process is illustrated in Part II of the proof of Theorem 3.4.6 directly below.
Theorem 3.4.6:
The Euclidean Parallel Postulate is equivalent to the Converse of the Alternate Interior Angle Theorem.
Proof:
[ Part I:
Prove: If EPP is true, then the Converse of the A I A Theorem is true. ]
Assume that the Euclidean Parallel Postulate is true.
Let
l
and
m
be two parallel lines and let
t
be a transversal
intersecting
l
and
m
at points A and B, respectively.
[ We NTS that the alternate interior angles formed by
t
at points A
and
B
are congruent
.
]
Let
1
and
2
be two alternate interior angles at A and B,
respectively.
Let point C on
l
and point D on
m
be positioned such
that
1
BAC
and
2
ABD
.
[We show that
12
using proofbycontradiction.]
Suppose, by way of contradiction, that
and
are not congruent.
Therefore, m(
1
)
m(
2
) .
By Postulate 12 (The Angle Construction Postulate), there is a unique ray
BE
in the halfplane side of
transversal
t
that contains point D such that
m(
ABE
)
=
m(
1
) . Let line
n
=
BE
and note that
line
m
=
BD
.
Since m(
)
m(
1
) , m(
)
m(
) .
Therefore,
rays
BD
and
BE
are two distinct rays, and so,
m
and
n
are two distinct
lines.
Now, line
t
is a transversal of lines
n
and
l
which forms
alternate interior angles
1
and
at
A
and B, resp.
Also,
1
by definition.
Therefore, by the Alternate
Interior Angle Theorem (Theorem 3.4.1),
lines
l
and
n
are
parallel .
Thus, lines
m
and
n
are two distinct lines, both
containing B and parallel to line
l , which contradicts the EPP
(which was assumed to be true).
Therefore,
.
Therefore, if the EPP is true, then the Converse of the Alternate Interior Angle Theorem is true. [ End of Part I ]
l
m
B
A
2
1
t
m
l
D
C
m(
ABE)
=
m(
1 )
l
m
n
B
A
2
1
t
m
l
E
D
C
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[ Part II:
Prove: If the Converse of the A I A Theorem is true, then the EPP is true. ]
Suppose that the statement of the Converse of the Alternate Interior Angle Theorem is true.
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 Spring '10
 Shirley
 Geometry

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