Step-by-step explanation:
Math 444/445 Geometry for Teachers Summer 2008
Supplement A: Rays, Angles, and Betweenness
This handout is meant to be read in place of Sections 5.6–5.7 in Venema’s text [V]. You should read
these pages after reading Venema’s Section 5.5.
Betweenness of Points
By definition, a point B is between two other points A and C if all three points are collinear and AB +BC =
AC. Although this definition is unambiguous and easy to state, it is not always easy to work with in proofs,
because we may not always know what the distances AB, BC, and AC are.
There are other ways of thinking about betweenness of points using coordinate functions, which are often
much more useful. Before stating them, let us establish some terminology.
We say that two or more mathematical statements are equivalent if any one of them implies all the
others. For example, if P and Q are mathematical statements, then to say they are equivalent is to say that
P implies Q and Q implies P, or to put it another way, “P if and only if Q.” Given four statements P, Q,
R, and S, if we wish to prove them all equivalent, we don’t have to prove that every statement implies all
of the others; for example, it would suffice to prove that P ⇒ Q ⇒ R ⇒ S ⇒ P (that is, P ⇒ Q, Q ⇒ R,
R ⇒ S, and S ⇒ P), for then if any one of the four statements is true, we can combine these implications
to arrive at any other.
If x, y, z are three real numbers, we say that y is between x and z if either x < y < z or x > y > z.
Theorem A.1 (Betweenness Theorem for Points). Suppose A, B, and C are distinct points all lying
on a single line `. Then the following statements are equivalent:
(a) AB + BC = AC (i.e., A ∗ B ∗ C).
(b) B lies in the interior of the line segment AC.
(c) B lies on the ray
−→AC and AB < AC.
(d) For any coordinate function f : ` → R, the coordinate f(B) is between f(A) and f(C).
Proof. We will prove (a) ⇔ (b), (a) ⇔ (c), and (a) ⇔ (d).
The equivalence of (a) and (b) is just another way of restating the definition of interior points of a
segment:
B is an interior point of AC ⇔ B ∈ AC and B 6= A and B 6= C
⇔ A ∗ B ∗ C
⇔ AB + BC = AC.
Next, we will prove (a) ⇒ (c). Assuming A ∗ B ∗ C, we conclude that B ∈ AC by definition of segments,
and therefore B ∈
−→AC by definition of rays. The fact that AB + BC = AC implies by algebra that
AB = AC − BC, which is strictly less than AC because BC > 0.
Now we will prove (c) ⇒ (a). Assuming (c), the fact that B ∈
−→AC means by definition that either
B ∈ AC or A ∗ C ∗ B. If the latter is true, then AC + CB = AB. But then our assumption that AB < AC
implies AC + CB < AC, and subtracting AC from both sides we conclude that CB < 0, a contradiction.
So the only remaining possibility is that B ∈ AC. Since B is not equal to A or C, we must have A ∗ B ∗ C.
The next step is to prove that (a) ⇒ (d). Suppose AB + BC = AC, and let f : ` → R be any coordinate
function for `. For convenience, write a = f(A), b = f(B), and c = f(C), so that AB = |b−a|, AC = |c−a|,
and BC = |c − b| by the Ruler Postulate. Our assumption then becomes the following equation:
|b − a| + |c − b| = |c − a|. (A.1)
1
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Answers & Comments
Step-by-step explanation:
Math 444/445 Geometry for Teachers Summer 2008
Supplement A: Rays, Angles, and Betweenness
This handout is meant to be read in place of Sections 5.6–5.7 in Venema’s text [V]. You should read
these pages after reading Venema’s Section 5.5.
Betweenness of Points
By definition, a point B is between two other points A and C if all three points are collinear and AB +BC =
AC. Although this definition is unambiguous and easy to state, it is not always easy to work with in proofs,
because we may not always know what the distances AB, BC, and AC are.
There are other ways of thinking about betweenness of points using coordinate functions, which are often
much more useful. Before stating them, let us establish some terminology.
We say that two or more mathematical statements are equivalent if any one of them implies all the
others. For example, if P and Q are mathematical statements, then to say they are equivalent is to say that
P implies Q and Q implies P, or to put it another way, “P if and only if Q.” Given four statements P, Q,
R, and S, if we wish to prove them all equivalent, we don’t have to prove that every statement implies all
of the others; for example, it would suffice to prove that P ⇒ Q ⇒ R ⇒ S ⇒ P (that is, P ⇒ Q, Q ⇒ R,
R ⇒ S, and S ⇒ P), for then if any one of the four statements is true, we can combine these implications
to arrive at any other.
If x, y, z are three real numbers, we say that y is between x and z if either x < y < z or x > y > z.
Theorem A.1 (Betweenness Theorem for Points). Suppose A, B, and C are distinct points all lying
on a single line `. Then the following statements are equivalent:
(a) AB + BC = AC (i.e., A ∗ B ∗ C).
(b) B lies in the interior of the line segment AC.
(c) B lies on the ray
−→AC and AB < AC.
(d) For any coordinate function f : ` → R, the coordinate f(B) is between f(A) and f(C).
Proof. We will prove (a) ⇔ (b), (a) ⇔ (c), and (a) ⇔ (d).
The equivalence of (a) and (b) is just another way of restating the definition of interior points of a
segment:
B is an interior point of AC ⇔ B ∈ AC and B 6= A and B 6= C
⇔ A ∗ B ∗ C
⇔ AB + BC = AC.
Next, we will prove (a) ⇒ (c). Assuming A ∗ B ∗ C, we conclude that B ∈ AC by definition of segments,
and therefore B ∈
−→AC by definition of rays. The fact that AB + BC = AC implies by algebra that
AB = AC − BC, which is strictly less than AC because BC > 0.
Now we will prove (c) ⇒ (a). Assuming (c), the fact that B ∈
−→AC means by definition that either
B ∈ AC or A ∗ C ∗ B. If the latter is true, then AC + CB = AB. But then our assumption that AB < AC
implies AC + CB < AC, and subtracting AC from both sides we conclude that CB < 0, a contradiction.
So the only remaining possibility is that B ∈ AC. Since B is not equal to A or C, we must have A ∗ B ∗ C.
The next step is to prove that (a) ⇒ (d). Suppose AB + BC = AC, and let f : ` → R be any coordinate
function for `. For convenience, write a = f(A), b = f(B), and c = f(C), so that AB = |b−a|, AC = |c−a|,
and BC = |c − b| by the Ruler Postulate. Our assumption then becomes the following equation:
|b − a| + |c − b| = |c − a|. (A.1)
1