4.
Measures of segments and angles
It is clear that segment congruence is an equivalence relation on the set of all
segments. For each segment AB we may use [AB] to denote the congruence
equivalence class of AB. Recall that line AB is totally ordered. We may view AB as a
segment together with an order such that A ≺ B. For any segment CD, we can construct
a new segment AP on the ray r(A, B) by laying off CD on the ray r(B, B'), where A ∗ B ∗
B', such that BB' ≅ CD. We write this new segment AB' as AB + CD.
Note that CD + AB is a segment on ray r(C, D) with endpoint C. Of course, AB + CD
and CD + AB are certainly distinct segments. However, AB + CD ≅ CD + AB. For the
congruence equivalence classes [AB], [CD], we define [AB] + [CD] := [AB + CD].
If EF is a segment, it is easy to see that
(AB + CD) + EF ≅ AB + (CD + EF).
We then have
[AB] + [CD] = [CD] + [AB].
([AB] + [CD]) + [EF] = [AB] + ([CD] + [EF]).
Let 1/2 • AB denote the segment whose endpoints are A and the midpoint of AB. For
each positive integer k, let k · AB denote the segment obtained by laying off k copies of
AB on the ray r(A, B) starting from A; let us define the segment,
For integers p ∈ Z and positive integer q ∈ Z+, we write,
Theorem 2.1. Fix a segment OI, called unit segment. There exists a unique mapping
from the set of all segments to the set R+ of positive real numbers, AB 7→ |AB|,
satisfying the properties:
(a) |OI| = 1.
(b) |AB| = |CD| if and only if AB ≅ CD.
(c) A ∗ B ∗ C if and only if |AC| = |AB| + |BC|.
(d) |AB| < |CD| if and only if AB < CD.
(e) For each positive real number a, there exists a segment AB such that |AB| = a.
Proof. (Sketch) Fix open ray r°(O, I). Each segment is congruent to a segment OA with
unique point A ∈ r°(O, I). It suffices to assign for each point A ∈ r°(O, I) a positive real
number. The right endpoints of the segments 2^p q • OI are assigned to numbers 2^p q,
known as dyatic rational numbers. If Archimedes’ axiom is satisfied, then every point
in r°(O, I) has a decimal expression with base 2, and the point is assigned to the real
number with decimal expression.
Given a real number a > 0. Let Σ1 := {P ∈ r°(O, I) : |OQ| ≤ a} and Σ2 := r°(O, I) \ Σ1.
Then {Σ1, Σ2} is a Dedekind cut of r°(O, I). There exists a unique Q such that Σ1 = [O,
Q] and Σ2 = (Q, •). We must have |OQ| = a.
Definition 1.
(a) Two angles ∠AOB and ∠A'O'B' are said to be addable if there exists
an angle ∠AOC such that B ∈ ˚∠AOC and ∠BOC ≅ ∠A'O'B'. We define the partial
addition ∠AOB + ∠B'O'C' := ∠AOC.
(b) An half-plane is also known as a flat angle. We assume that all flat angles are
congruent.
Theorem 2.2 (Degree measure of angles). There exists a unique mapping from the set
of all angles to the interval (0, 180) of real numbers, ∠A 7→ ∠A° , satisfying the
properties:
(a) ∠A° = 90° if ∠A is a right angle.
(b) ∠A° = ∠B° if and only if ∠A ≅ ∠B.
(c) If h(O, B) ≅ ∠oAOC, then ∠AOC° = ∠AOB° + ∠BOC°.
(d) ∠A° < ∠B° if and only if ∠A < ∠B.
(e) If ∠A, ∠B are complementary, then ∠A° + ∠B° = 180°.
(f ) For each positive real number a ∈ (0, 180), there exists an angle ∠AOB such
that
∠AOB° = 180°.
Proof. (Sketch) Fix ray r(O, A) and consider angles ∠AOB. Since each angle can be
bisected, we denote by 1/2 · ∠AOB the angle ∠AOP, where r(O, P) is the bisector of
∠AOB. We then have angles,
If ∠AOB is addable to (k − 1)∠AOB , we define k∠AOB := (k − 1)∠AOB + ∠AOB.
We thus have,
as long as they are addable to each other. Let [∠AOB] denote the equivalence class of
angles congruent to ∠AOB. Then
Now we fix a right angle ∠ROT. Assign the angles 2^−pq ∠ROT to the real
numbers 90 • 2^−pq, that is,
Dedekind’s axiom ensures that the assignment can be extended into a bijection from
angle congruence classes onto the open interval (0, 180) of real numbers expressed in
decimal form of base 2, satisfying (a)-(f). For instance, for part (e) about complementary
angles ∠A, ∠B, we see that 1/2∠A + 1/2∠B is a right angle so that
Corollary 2.3 (Consecutive Interior Angle Theorem). If two distinct lines cut by a
transversal have a pair of consecutive interior angles whose angle sum is a flat angle,
then the two lines are parallel.
Proof. We assume in Figure 1 that ∠AP Q + ∠BQP is congruent to a flat angle. Since
∠B'QP + ∠BQP is congruent is a flat angle, we see that ∠AP Q ≅ ∠B'QP. So m||l.
Proposition 2.4 (Triangle inequality). Let A, B, C be three distinct non-collinear points.
Then
|AC| < |AB| + |BC|.
Proof. There exists a unique point D such that A ∗ B ∗ D and BD ≅= BC by Congruence
Axiom 1. See Figure 9. Then ∠BCD ≅ BDC because of the isosceles triangle ∆BCD.
Since ∠ACD > ∠BCD ≅ ∠BDC, we have ∠ACD > ∠BDC = ∠ADC. Hence AD > AC
by the angle-opposite-side relation. Since AD ≅ AB + BC, we obtain AB + BC > AC.
Subsequently, |AB| + |BC| > |AC| by measure of segments.
Proposition 2.5 (Included angle-opposite-side relation of equal sides). Given triangles
∆ABC, ∆A'B'C' and AB ≅ A'B', AC ≅ A'C' . Then ∠A > ∠A' if and only if BC > B'C'.
Proof. “⇒”: Let ∠BAC > ∠B'A'C' . Draw a ray r(A, P) between r(A, B) and r(A, C) such
that ∠BAP ≅ ∠B'A' Let D be a point on r(A, P) such that AD ≅ A'C' , and E the
intersection of r(A, P) and BC such that B ∗ E ∗ C. Draw the angle bisector of ∠CAD
and its intersection with CE at F between C and E. Then ∆ACF ≅ ∆ADF by SAS. Hence
FC ≅ FD. Note that BF + FD > BD by triangle inequality. Then BF +F C > BD. Since BF
+ F C ≅ BC and B'C' ≅ BD. We obtain BC > B'C' .“⇐”: Let BC > B'C'. If ∠BAC ≅ ∠B'A'C'
; then ∆ABC ≅ ∆A'B'C' by SAS; so BC ≅ B'C', which is a contradiction. If ∠BAC <
∠B'A'C', then B'C' > BC by what just proved previously, which is also a contradiction.
Hence we have ∠BAC > ∠B'A'C' by trichotomy of angles.
ASSESSMENT:
1. If m∠RST = (12x -1)°, m∠RSU = (9x -15)°, and m∠UST = 53°, find each measure.
2. In the diagram to the right, is ∆ABC ≅ ∆CDA? Answer yes or no and breifly explain
your reasoning.
REFERENCE:
UniversityofNorthGeorgia
http://blog.ung.edu
UniversityofKentucky
https://www.ms.uky.edu
Elephango
https://www.elephango.com
Virtual Nerd
https://virtualnerd.com
Sites.google.com
https://sites.google.com
Youtube:
https://www.youtube.com/watch?v=qUx
https://youtu.be/gmrSp2LNKno
