Download Tutorial Note 8

MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Tutorial Note 8
Apr 11, 2012 (Week 10)
MATH4221 Euclidean and Non-Euclidean Geometries
Tutorial Note 8
Topics covered in week 9:
12. Angle Sum of Triangles
13. Equivalent Statements of the Euclidean Parallel Postulate
12. Angle Sum of Triangles
What you need to know:
 Saccheri-Legendre Theorem:


Angle sum of every triangle ≤ 180°
Euclid’s Fifth Postulate ⇔ Euclidean Parallel Postulate
⇔ Angle sum of some triangle = 180° ⇔ Angle sum of every triangle = 180°
Defect of a triangle
Saccheri-Legendre Theorem is an important theorem in this course. It states that in neutral
geometry, the angle sum of a triangle is always less than or congruent to 𝟏𝟖𝟎°.
Remark:
We emphasize once again that we do not have the notion of “angle measure” in this
course. The above statement is just saying that any angle of a triangle is less than
or congruent to the supplementary angle of the sum of the two remaining angles.
Example 12.1 Prove that he exterior angle of a triangle is greater than or congruent to the sum
of its two interior opposite angles.
Proof:
Let Δ𝐴𝐵𝐶 be an arbitrary triangle.
Without loss of generality, we consider the exterior angle ∠𝐴𝐶𝐷 of Δ𝐴𝐵𝐶, where 𝐷 is a point
such that 𝐵 ∗ 𝐶 ∗ 𝐷. Then,
(𝐵 ∗ 𝐶 ∗ 𝐷)
∠𝐴𝐶𝐷 ≅ 180° − ∠𝐴𝐶𝐵
(Saccheri-Legendre Theorem, Δ𝐴𝐵𝐶)
≥ ∠𝐴𝐵𝐶 + ∠𝐵𝐴𝐶
∎
Page 1 of 10
MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Tutorial Note 8
Apr 11, 2012 (Week 10)
We now distinguish clearly between “Euclid’s Fifth Postulate” and “Euclidean Parallel Postulate”.
Euclid’s Fifth Postulate states that two lines having a sum of interior angles on the same side of
a transversal less than 180° must meet, whereas Euclidean Parallel Postulate is our familiar
“unique parallel line” statement. These two statements are equivalent, and they are also
equivalent to the statement that any triangle has an angle sum congruent to 180°.
Let’s look at another equivalent statement of the Parallel Postulate regarding the angle sum of a
triangle.
Example 12.2 (Trudeau 4.4)
The following is a “proof” of the statement that the angle sum
of any triangle is 180° in neutral geometry. What is wrong with this proof?
“Proof”:
We pretend that we do not know what the angle sum of a triangle is, and we let
𝑥 be that congruence class of angles.
Now let Δ𝐴𝐵𝐶 be any triangle, and 𝐷 be a point such that 𝐵 ∗ 𝐷 ∗ 𝐶, so that
𝐷 is in the interior of ∠𝐵𝐴𝐶.
Then considering the angle sums of Δ𝐴𝐵𝐷 and Δ𝐴𝐶𝐷, we have
∠𝐴𝐵𝐷 + ∠𝐵𝐴𝐷 + ∠𝐴𝐷𝐵 ≅ 𝑥, and
∠𝐴𝐶𝐷 + ∠𝐶𝐴𝐷 + ∠𝐴𝐷𝐶 ≅ 𝑥.
Adding up we get
∠𝐴𝐵𝐷 + ∠𝐵𝐴𝐷 + ∠𝐴𝐷𝐵 + ∠𝐴𝐶𝐷 + ∠𝐶𝐴𝐷 + ∠𝐴𝐷𝐶 ≅ 2𝑥
∠𝐴𝐵𝐷 + (∠𝐵𝐴𝐷 + ∠𝐶𝐴𝐷) + ∠𝐴𝐶𝐷 + (∠𝐴𝐷𝐵 + ∠𝐴𝐷𝐶) ≅ 2𝑥
(∠𝐴𝐵𝐶 + ∠𝐵𝐴𝐶 + ∠𝐴𝐶𝐵) + 180° ≅ 2𝑥.
But being the angle sum of Δ𝐴𝐵𝐶, we have ∠𝐴𝐵𝐶 + ∠𝐵𝐴𝐶 + ∠𝐴𝐶𝐵 ≅ 𝑥.
Therefore 𝑥 ≅ 180°.
∎
Solution:
The flaw is in the very first statement. It is not valid to simply assume that “all triangles have
congruent angle sums”.
In fact this unstated assumption is equivalent to the Euclidean Parallel Postulate:
(⇒) From the above proof, this assumption implies that the angle sum of any triangle is
congruent to 180°, which implies the Euclidean Parallel Postulate (Theorem 4.9 in the
Lecture Note).
(⇐) The Parallel Postulate implies that all triangles have angle sum congruent to 180°, so all
triangles have congruent angle sums.
Page 2 of 10
MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Tutorial Note 8
Apr 11, 2012 (Week 10)
The defect of a triangle Δ𝐴𝐵𝐶, denoted as 𝛿Δ𝐴𝐵𝐶, is defined to be an equivalence class of
angles congruent to 𝟏𝟖𝟎° minus its angle sum. One can treat the defect of a triangle just like
its area in the usual sense, because it satisfies the following Lebesgue measure properties:
(i) (“Positivity”) Every triangle has an angle sum less than or congruent to 180°, i.e. a
well-defined defect.
(ii) (Congruential Invariance) Congruent triangles have congruent defects.
(iii) (Additivity) If a triangle is partitioned into two smaller triangles, then the sum of defects
of these two smaller triangles is congruent to the defect of the original triangle.
Example 12.3 (Prenowitz & Jordan 4.II.3, 4) Let Δ𝐴𝐵𝐶 be a triangle. Show that
(a) If 𝑃 is a point in the interior of Δ𝐴𝐵𝐶, then we have
𝛿Δ𝐴𝐵𝐶 ≅ 𝛿Δ𝑃𝐴𝐵 + 𝛿Δ𝑃𝐵𝐶 + 𝛿Δ𝑃𝐴𝐶.
(b) If 𝐴 ∗ 𝑃 ∗ 𝐵, 𝐵 ∗ 𝑄 ∗ 𝐶 and 𝐴 ∗ 𝑅 ∗ 𝐶, then we have
𝛿Δ𝐴𝐵𝐶 ≅ 𝛿Δ𝐴𝑃𝑅 + 𝛿Δ𝐵𝑄𝑃 + 𝛿Δ𝐶𝑅𝑄 + 𝛿Δ𝑃𝑄𝑅.
Proof:
[We aim to divide Δ𝐴𝐵𝐶 into smaller triangles step-by-step, so as to apply Lemma 3.34.]
(a) By Crossbar Theorem, ⃗⃗⃗⃗⃗
𝐴𝑃 meets 𝐵𝐶 at some point 𝐷. Since 𝑃 is in the interior of
⃗⃗⃗⃗⃗ ∗ 𝐵𝑃
⃗⃗⃗⃗⃗ ∗ 𝐵𝐷
⃗⃗⃗⃗⃗⃗ and so 𝐴 ∗ 𝑃 ∗ 𝐷. Therefore we have
∠𝐴𝐵𝐶 = ∠𝐴𝐵𝐷, it follows that 𝐵𝐴
(𝐵 ∗ 𝐷 ∗ 𝐶)
𝛿Δ𝐴𝐵𝐶 ≅ 𝛿Δ𝐴𝐵𝐷 + 𝛿Δ𝐴𝐶𝐷
(𝐴 ∗ 𝑃 ∗ 𝐷)
≅ (𝛿Δ𝑃𝐴𝐵 + 𝛿Δ𝑃𝐷𝐵) + (𝛿Δ𝑃𝐴𝐶 + 𝛿Δ𝑃𝐷𝐶)
≅ 𝛿Δ𝑃𝐴𝐵 + (𝛿Δ𝑃𝐷𝐵 + 𝛿Δ𝑃𝐷𝐶) + 𝛿Δ𝑃𝐴𝐶
(𝐵 ∗ 𝐷 ∗ 𝐶)
≅ 𝛿Δ𝑃𝐴𝐵 + 𝛿Δ𝑃𝐵𝐶 + 𝛿Δ𝑃𝐴𝐶
𝐴
𝑃
𝑃
𝐵
𝐷
∎
𝐴
𝐵
𝐶
𝑂
𝑄
𝑅
𝐶
⃗⃗⃗⃗⃗ ∗ 𝐵𝑅
⃗⃗⃗⃗⃗ , so
(b) By Crossbar Theorem, ⃗⃗⃗⃗⃗
𝐴𝑄 meets 𝑃𝑅 at some point 𝑂. This gives ⃗⃗⃗⃗⃗
𝐵𝑃 ∗ 𝐵𝑂
⃗⃗⃗⃗⃗ ∗ 𝐵𝑂
⃗⃗⃗⃗⃗ ∗ 𝐵𝑅
⃗⃗⃗⃗⃗ ∗ 𝐵𝑅
⃗⃗⃗⃗⃗ , so 𝐵𝐴
⃗⃗⃗⃗⃗ ∗ 𝐵𝑂
⃗⃗⃗⃗⃗ ∗ 𝐵𝐶
⃗⃗⃗⃗⃗ .
⃗⃗⃗⃗⃗ . But 𝐴 ∗ 𝑅 ∗ 𝐶 implies that 𝐵𝐴
⃗⃗⃗⃗⃗ ∗ 𝐵𝐶
𝐵𝐴
⃗⃗⃗⃗⃗ ∗ 𝐵𝑄
⃗⃗⃗⃗⃗ , and 𝐴 ∗ 𝑂 ∗ 𝑄. Therefore we have
Thus ⃗⃗⃗⃗⃗
𝐵𝐴 ∗ 𝐵𝑂
(𝐵 ∗ 𝑄 ∗ 𝐶)
𝛿Δ𝐴𝐵𝐶 ≅ 𝛿Δ𝐴𝐵𝑄 + 𝛿Δ𝐴𝐶𝑄
(𝐴 ∗ 𝑃 ∗ 𝐵 and 𝐴 ∗ 𝑅 ∗ 𝐶)
≅ (𝛿Δ𝐴𝑄𝑃 + 𝛿Δ𝐵𝑄𝑃) + (𝛿Δ𝐴𝑅𝑄 + 𝛿Δ𝐶𝑅𝑄)
≅ (𝛿Δ𝐴𝑃𝑂 + 𝛿Δ𝑄𝑃𝑂) + 𝛿Δ𝐵𝑄𝑃 + (𝛿Δ𝐴𝑅𝑂 + 𝛿Δ𝑄𝑅𝑂) + 𝛿Δ𝐶𝑅𝑄 (𝐴 ∗ 𝑂 ∗ 𝑄)
≅ (𝛿Δ𝐴𝑃𝑂 + 𝛿Δ𝐴𝑅𝑂) + 𝛿Δ𝐵𝑄𝑃 + (𝛿Δ𝑄𝑃𝑂 + 𝛿Δ𝑄𝑅𝑂) + 𝛿Δ𝐶𝑅𝑄
(𝑃 ∗ 𝑂 ∗ 𝑅)
≅ 𝛿Δ𝐴𝑃𝑅 + 𝛿Δ𝐵𝑄𝑃 + 𝛿Δ𝐶𝑅𝑄 + 𝛿Δ𝑃𝑄𝑅
∎
Page 3 of 10
MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Tutorial Note 8
Apr 11, 2012 (Week 10)
13. Equivalent Statements of the Euclidean Parallel Postulate
What you need to know:
 Definition of similar triangles
 Definition of Saccheri quadrilaterals, Lambert quadrilaterals, rectangles
 The following are equivalent statements:
 Euclid’s Fifth Postulate
 Euclidean Parallel Postulate
 Converse of Alternate Angle Theorem
 Angle sum of a triangle = 𝟏𝟖𝟎°
 Wallis’ Postulate


Existence of a rectangle
Legendre’s Postulate
This section is mainly about the history and development of geometry regarding parallelism.
The main focus is to be aware of the above equivalent statements of the Euclidean Parallel
Postulate.
The following examples help you to get familiar with the notions of a
Saccheri quadrilateral and a Lambert quadrilateral, which are tools for
studying parallelism used by the mathematicians Saccheri and Lambert.
A Saccheri quadrilateral □𝐴𝐵𝐶𝐷 is a quadrilateral in which there are two
adjacent right angles, say ∠𝐷𝐴𝐵 and ∠𝐴𝐵𝐶, and the pair of opposite
sides 𝐴𝐷 and 𝐵𝐶 are congruent. Names are given to the sides and
angles to a Saccheri quadrilateral as shown in the figure on the right.
Summit angles
𝐷
𝐶
Summit
Side
Side
𝐴
Base
𝐵
Saccheri quadrilateral
Example 13.1 Show that in any Saccheri quadrilateral, the summit is always greater than or
equal to the base.
Proof:
Let □𝐴𝐵𝐶𝐷 be a Saccheri quadrilateral with base 𝐴𝐵. We aim to show that 𝐶𝐷 ≥ 𝐴𝐵.
In the right triangle Δ𝐴𝐵𝐷, we have ∠𝐴𝐵𝐷 + ∠𝐴𝐷𝐵 + ∠𝐵𝐴𝐷 ≤ 180° by Saccheri-Legendre
Theorem. Since ∠𝐵𝐴𝐷 ≅ 90°, we have
∠𝐴𝐵𝐷 + ∠𝐴𝐷𝐵 ≤ 90°.
𝐷
𝐶
𝐴
𝐵
Page 4 of 10
MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Tutorial Note 8
Apr 11, 2012 (Week 10)
We will see later in the lecture that the opposite sides of a Saccheri quadrilateral are parallel,
which implies that 𝐷 must be in the interior of ∠𝐴𝐵𝐶. Thus,
∠𝐶𝐵𝐷 + 90° ≥ ∠𝐶𝐵𝐷 + ∠𝐴𝐵𝐷 + ∠𝐴𝐷𝐵
≅ ∠𝐴𝐵𝐶 + ∠𝐴𝐷𝐵
≅ 90° + ∠𝐴𝐷𝐵,
so ∠𝐴𝐷𝐵 ≤ ∠𝐶𝐵𝐷.
Since we also have 𝐴𝐷 ≅ 𝐶𝐵 and 𝐵𝐷 = 𝐷𝐵 , applying the Angle-side Correspondence
Theorem to Δ𝐴𝐷𝐵 and Δ𝐶𝐵𝐷, we have 𝐶𝐷 ≥ 𝐴𝐵.
∎
We have seen that a greater side of a triangle corresponds to a greater angle, and that isosceles
triangles have congruent base angles. This follows analogously for quadrilaterals with two
adjacent right angles. A greater side of such a quadrilateral corresponds to a greater summit
angle, and Saccheri quadrilaterals have congruent summit angles.
Example 13.2 (Angle-side Correspondence Theorem)
Let □𝐴𝐵𝐶𝐷 be a quadrilateral with
∠𝐷𝐴𝐵 ≅ ∠𝐴𝐵𝐶 ≅ 90°. Show that 𝐴𝐷 < 𝐵𝐶 if and only if ∠𝐵𝐶𝐷 < ∠𝐴𝐷𝐶.
Proof:
(⇒) If 𝐴𝐷 < 𝐵𝐶, then by Segment Relocation Axiom, there exists 𝐸 such that 𝐵 ∗ 𝐸 ∗ 𝐶 and
𝐵𝐸 ≅ 𝐴𝐷. So □𝐴𝐵𝐸𝐷 is a Saccheri quadrilateral.
𝐶
Since 𝐵 is in the interior of ∠𝐴𝐷𝐸 and 𝐸 is in the interior of ∠𝐵𝐷𝐶,
𝐷
𝐸
it follows that 𝐸 is in the interior of ∠𝐴𝐷𝐶, so ∠𝐴𝐷𝐸 < ∠𝐴𝐷𝐶.
On the other hand, since □𝐴𝐵𝐸𝐷 is a Saccheri quadrilateral, we have
𝐵
∠𝐴𝐷𝐸 ≅ ∠𝐵𝐸𝐷 (Summit angles of Saccheri quadrilateral) 𝐴
> ∠𝐸𝐶𝐷 (Exterior Angle Theorem, Δ𝐶𝐷𝐸)
= ∠𝐵𝐶𝐷 (𝐵 ∗ 𝐸 ∗ 𝐶)
so ∠𝐵𝐶𝐷 < ∠𝐴𝐷𝐶.
(⇐) If ∠𝐵𝐶𝐷 < ∠𝐴𝐷𝐶, then by trichotomy of “<” we can only have 𝐴𝐷 < 𝐵𝐶, 𝐴𝐷 ≅ 𝐵𝐶 or
𝐴𝐷 > 𝐵𝐶:
(i) If 𝐴𝐷 > 𝐵𝐶, then by the same argument in “⇒” we have ∠𝐴𝐷𝐶 < ∠𝐵𝐶𝐷, which is a
contradiction.
(ii) If 𝐴𝐷 ≅ 𝐵𝐶, then □𝐴𝐵𝐶𝐷 is a Saccheri quadrilateral, so ∠𝐴𝐷𝐶 ≅ ∠𝐵𝐶𝐷, which is
again a contradiction.
So we must have 𝐴𝐷 < 𝐵𝐶.
∎
Page 5 of 10
MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Tutorial Note 8
Apr 11, 2012 (Week 10)
Exercise
1. (Greenberg 4.15)
(a) Show that given any triangle Δ𝐴𝐵𝐶, there exists Δ𝐴′ 𝐵 ′ 𝐶 ′ having congruent angle sum as
Δ𝐴𝐵𝐶 and 2∠𝐵 ′ 𝐴′ 𝐶 ′ ≤ ∠𝐵𝐴𝐶.
Remark:
“2∠𝐵 ′ 𝐴′ 𝐶 ′ ” is just a short-hand notation for “∠𝐵 ′ 𝐴′ 𝐶 ′ + ∠𝐵 ′ 𝐴′ 𝐶 ′ ”. We have
not defined “scalar multiplication” on equivalent classes of angles and of
segments.
(b) Hence, give an alternative proof to Saccheri-Legendre’s Theorem by contradiction.
Hint:
You may also use the proven fact that the sum of any two angles in a triangle is
smaller than 180°.
2. (O’Leary 10.4.12) (Angle in semicircle) Show that the following statement is equivalent to
the Parallel Postulate: There exists four distinct points 𝑂, 𝐴, 𝐵 and 𝐶 such that 𝐴 ∗ 𝑂 ∗ 𝐵,
𝑂𝐴 ≅ 𝑂𝐵 ≅ 𝑂𝐶 and ∠𝐴𝐶𝐵 is a right angle.
Hint:
Consider the angle sum of a triangle.
3. (Wallace & West 3.6.13, Prenowitz & Jordan 3.II.1, 2, 3.III.2 adapted) Let Δ𝐴𝐵𝐶 be a triangle
and 𝑀, 𝑁 be the mid-points of the sides 𝐴𝐵 and 𝐴𝐶 respectively.
(a) Show that if 𝑄 and 𝑅 are respectively the feet of perpendicular of 𝐵 and 𝐶 on ⃡⃗⃗⃗⃗⃗
𝑀𝑁,
then □𝐵𝑄𝑅𝐶 is a Saccheri quadrilateral. Also show that the sum of the summit angles
∠𝑄𝐵𝐶 and ∠𝑅𝐶𝐵 is congruent to the angle sum of Δ𝐴𝐵𝐶.
⃡⃗⃗⃗⃗ and that 2𝑀𝑁 ≤ 𝐵𝐶.
(b) Hence, show that ⃡⃗⃗⃗⃗⃗
𝑀𝑁 is parallel to 𝐵𝐶
(c) The Mid-point Theorem in high-school geometry states that if Δ𝐴𝐵𝐶 is a triangle and 𝑀,
𝑁 are the mid-points of the sides 𝐴𝐵 and 𝐴𝐶 respectively, then 2𝑀𝑁 ≅ 𝐵𝐶. Using the
above results, show that the Mid-point Theorem is equivalent to Euclid’s Fifth Postulate.
4. (Prenowitz & Jordan 4.III.1, 2) The following exercises are about defects of a triangle.
(a) Let Δ𝐴𝐵𝐶 be a right triangle. Show that there exists a triangle Δ𝐴′ 𝐵 ′ 𝐶 ′ such that
𝛿Δ𝐴′ 𝐵 ′ 𝐶 ′ ≅ 2𝛿Δ𝐴𝐵𝐶.
(b) Let Δ𝐴𝐵𝐶 be a triangle (not necessarily a right triangle). Show that there exists a triangle
Δ𝐴′ 𝐵 ′ 𝐶 ′ such that 2𝛿Δ𝐴′ 𝐵 ′ 𝐶 ′ ≅ 𝛿Δ𝐴𝐵𝐶.
Hint:
The construction of a Saccheri quadrilateral as in Problem 3(a) may be helpful.
Page 6 of 10