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Subsections
We remark that the point of the axiom is not the existence of the
parallel, but the uniqueness. We will see below that existence
actually follows from what we already know.
It is sometimes convenient to think of a line as being parallel to itself,
so we make the following formal definition. Two lines are not parallel if
they have exactly one point in common; otherwise they are parallel.
Theorem 3.1
In the set of all lines in the plane, the relation of being parallel is an
equivalence relation.
We will meet the following situation
some number of times. We are given two lines
and
, and a third
line
, where
crosses
at
and
crosses
at
. Choose a point
on
, and choose a point
on
, where
and
lie on opposite sides of the line
. Then
and
are referred to as
alternate interior angles.
In any given situation, there are two distinct pairs of alternate interior
angles. That is, let
be some point on
, where
and
lie
on opposite sides of
, and let
be some point on
, where
and
lie on opposite sides of
. Then one could also regard
and
as being alternate interior
angles. However, observe that
and
. It follows that one pair of
alternate interior angles are equal if and only if the other pair of
alternate interior angles are equal.
Proposition 3.2
If the alternate interior angles are equal, then the lines
and
are parallel.
Proof.
Suppose not. Then the lines
![$ k_1$](img149.gif)
and
![$ k_2$](img150.gif)
meet at some point
![$ D$](img18.gif)
. If
necessary, we interchange the roles of the
![$ B_i$](img166.gif)
and the
![$ C_i$](img167.gif)
so that
![$ \angle B_1A_1A_2$](img157.gif)
is an exterior angle of
![$ \triangle A_1A_2D$](img168.gif)
. Then
![$ D$](img18.gif)
and
![$ B_2$](img156.gif)
lie on the same side of
![$ m$](img50.gif)
, so
![$ \angle DA_2A_1=\angle B_2A_2A_1$](img169.gif)
. By
the exterior angle inequality,
so we have reached a contradiction.
Let
be a line, and let
be a point not on
. Pick some point
on
and draw the line
through
and
. By Axiom 7, we can find a line
through
so that the alternate interior angles are equal. Hence we can find a
line through
parallel to
.
Theorem 3.3 (alternate interior angles equal)
Two lines
and
are parallel if and only if the alternate
interior angles are equal.
Proof.
To prove the forward direction, construct the line
![$ k_3$](img171.gif)
through
![$ A_2$](img152.gif)
,
where there is a point
![$ B_3$](img172.gif)
on
![$ k_3$](img171.gif)
, with
![$ B_3$](img172.gif)
and
![$ B_2$](img156.gif)
on the same
side of
![$ m$](img50.gif)
, so that
![$ m\angle B_3A_2A_1=m\angle B_1A_1A_2$](img173.gif)
. Then, by
Prop.
3.2,
![$ k_3$](img171.gif)
is a line through
![$ A_2$](img152.gif)
parallel to
![$ k_1$](img149.gif)
. Axiom
8 implies
![$ k_3=k_1$](img174.gif)
. Hence
![$ m\angle B_3A_2A_1=m\angle B_2A_2A_1$](img175.gif)
, and the desired conclusion follows.
The other direction is just Prop. 3.2,
restated as part of this theorem for convenience.
Theorem 3.4
The sum of the measures of the angles of a triangle is equal to
.
Proof.
Consider
![$ \triangle
ABC$](img31.gif)
, and let
![$ m$](img50.gif)
be the line passing through
![$ A$](img2.gif)
and
parallel to
![$ BC$](img33.gif)
. Let
![$ D$](img18.gif)
and
![$ E$](img15.gif)
be two points on
![$ m$](img50.gif)
, on opposite sides of
![$ A$](img2.gif)
, where
![$ D$](img18.gif)
and
![$ C$](img17.gif)
lie on opposite sides of the line
![$ AB$](img4.gif)
. Then
![$ B$](img3.gif)
and
![$ E$](img15.gif)
lie on opposite sides of
![$ AC$](img6.gif)
.
Exercise 3.1
Use alternate interior angles to complete the proof of this theorem.
A quadrilateral is a region bounded by four line segments, so it
has four verticies on its boundary.
Corollary 3.5
The sum of the measures of the angles of a quadrilateral is
.
Proof.
Cut the triangle into two triangles, and do the obvious computation.
A rectangle is a quadrilateral in which all four angles are right
angles.
Theorem 3.6
If
is a rectangle, then
is parallel to
, and
. Similarly,
is parallel to
and
.
Exercise 3.2
Prove this theorem.
- i.
- Prove that opposite pairs of sides are parallel.
- ii.
- Now cut the rectangle into two triangles; prove that these two
triangles are congruent. Conclude that opposite sides of the rectangle have
equal length.
Somewhat more generally, a parallelogram is a quadrilateral
in which opposite sides are parallel; that is,
is parallel to
, and
is parallel to
.
A rectangle with all four sides of equal length is a square; a
parallelogram with all four sides of equal length is a rhombus.
Exercise 3.3
Prove this theorem. (Hint: Draw a diagonal.)
Theorem 3.8
If
is a quadrilateral in which
and
, then
is a parallelogram.
Exercise 3.4
Prove this theorem.
Theorem 3.9
Let
be a parallelogram with diagonals of equal length (that is,
). Then
is a rectangle.
Exercise 3.5
Prove this theorem.
Next: 4 Lengths, areas and
Up: MAT 200 Course Notes on
Previous: 2 Triangles and congruence
Scott Sutherland
2002-12-18