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MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Tutorial Note 7
Mar 28, 2012 (Week 9)
MATH4221 Euclidean and Non-Euclidean Geometries
Tutorial Note 7
Topics covered in weeks 7-8:
11. Neutral Geometry
11. Neutral Geometry
What you need to know:
 Neutral geometry: Incidence + betweenness + congruence + continuity
 Alternate Angle Theorem: Converse does not hold in neutral geometry!



Two lines perpendicular to a common line are parallel
Exterior Angle Theorem
AAS Theorem and RHS Theorem for congruence of triangles

Existence and uniqueness of the mid-point, the angle bisector and the perpendicular
bisector
Addition operation (+) defined on congruence classes of segments and angles
The symbols “90°” and “180°”
Angle-side Correspondence Theorem (in one triangle and in two triangles)
Triangle Inequality




From now on, we will discard the Parallel Postulate from our list of axioms and study its
independence to the remaining 14 axioms. This subject is known as neutral geometry.
The Alternate Angle Theorem states that if the alternate angles of a pair of lines over a
transversal are congruent, then the pair of lines are parallel to each other.
Remark:
In secondary school you’ve also learnt the converse to this theorem, which states
that the alternate angles of a pair of parallel lines over any transversal are congruent.
Unfortunately this converse does not hold in our subject of neutral geometry – it is in
fact an equivalent statement of the Parallel Postulate.
Page 1 of 11
MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Example 11.1 (Noronha 1.5.1 adapted)
Tutorial Note 7
Mar 28, 2012 (Week 9)
A quadrilateral □𝐴𝐵𝐶𝐷 is an ordered collection of
four points such that no three of them are collinear and any pair of sides 𝐴𝐵,
𝐵𝐶, 𝐶𝐷 and 𝐷𝐴 either have no point in common or have only an endpoint in
common. A parallelogram is a quadrilateral whose opposite sides are parallel.
(a) Show that if opposite sides of a quadrilateral are congruent, then the
quadrilateral is a parallelogram.
(b) Show that if the diagonals of a quadrilateral bisect each other, then the
quadrilateral is a parallelogram.
Proof:
First let □𝐴𝐵𝐶𝐷 be a quadrilateral. In both parts we aim to show that ⃡𝐴𝐵 and ⃡𝐷𝐶 are
⃡ and 𝐵𝐶
⃡ are parallel.
parallel, and that 𝐴𝐷
𝐷
(a) We are given that 𝐴𝐵 ≅ 𝐶𝐷 and 𝐴𝐷 ≅ 𝐶𝐵. Together with
𝐵𝐷 ≅ 𝐷𝐵, we have Δ𝐴𝐵𝐷 ≅ Δ𝐶𝐷𝐵 by SSS Theorem.
𝐴
In particular, we have ∠𝐴𝐵𝐷 ≅ ∠𝐶𝐷𝐵 and ∠𝐴𝐷𝐵 ≅ ∠𝐶𝐵𝐷.
𝐵
Thus by Alternate Angle Theorem, we see that ⃡𝐴𝐵 and ⃡𝐷𝐶
⃡ and 𝐵𝐶
⃡ are parallel. So □𝐴𝐵𝐶𝐷 is a parallelogram.
are parallel, and 𝐴𝐷
(b) We are given that 𝐴𝐶 and 𝐵𝐷 meet at some point 𝑂 such
that 𝑂𝐴 ≅ 𝑂𝐶 and 𝑂𝐵 ≅ 𝑂𝐷.
Since ∠𝐴𝑂𝐵 and ∠𝐶𝑂𝐷 are vertically opposite angles, we
have ∠𝐴𝑂𝐵 ≅ ∠𝐶𝑂𝐷 (Example 9.1).
𝐶
∎
𝐷
𝐴
𝑂
𝐶
𝐵
Similarly, we also have ∠𝐴𝑂𝐷 ≅ ∠𝐶𝑂𝐵.
So by SAS Axiom, we have Δ𝐴𝑂𝐵 ≅ Δ𝐶𝑂𝐷 and Δ𝐴𝑂𝐷 ≅ Δ𝐶𝑂𝐵.
In particular, we have ∠𝐴𝐵𝑂 ≅ ∠𝐶𝐷𝑂 and ∠𝐴𝐷𝑂 ≅ ∠𝐶𝐵𝑂. But these just imply that
∠𝐴𝐵𝐷 ≅ ∠𝐶𝐷𝐵 and ∠𝐴𝐷𝐵 ≅ ∠𝐶𝐵𝐷.
⃡ and 𝐷𝐶
⃡ are parallel, and 𝐴𝐷
⃡ and 𝐵𝐶
⃡
Thus by Alternate Angle Theorem, we see that 𝐴𝐵
are parallel. So □𝐴𝐵𝐶𝐷 is a parallelogram.
Remark:
∎
The converses of the above statements also do not hold. In neutral geometry, there
exist parallelograms whose opposite sides are not congruent or whose diagonals do
not bisect each other.
A very useful corollary that can be immediately deduced from the Alternate Angle Theorem is
that two lines perpendicular to a common line are parallel. The converse is again not true in
neutral geometry – there can be a pair of parallel lines which do not have a common
perpendicular!
Page 2 of 11
MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Tutorial Note 7
Mar 28, 2012 (Week 9)
The Exterior Angle Theorem states that an exterior angle of a triangle is greater than both of its
remote interior angles.
Remark:
Don’t mix up this theorem with the statement that “an exterior angle of a triangle is
congruent to the sum of its two remote interior angles”. In neutral geometry, this
latter statement you learnt in secondary school is equivalent to the Parallel Postulate!
Example 11.2 It has been shown during the lecture, using Alternate Angle Theorem, that the
perpendicular to a line 𝑙 from a point 𝑃 not incident to 𝑙 is unique. Now
prove this statement using Exterior Angle Theorem instead.
Proof:
Suppose on the contrary that there exist two distinct perpendiculars
to 𝑙 from 𝑃.
Let the feet of perpendicular be 𝐴 and 𝐵 respectively, and let 𝐶
be a point incident to 𝑙 such that 𝐴 ∗ 𝐵 ∗ 𝐶 by Axiom (B2).
It follows that ∠𝑃𝐴𝐶 and ∠𝑃𝐵𝐶 are both right angles.
But by the Exterior Angle Theorem, we must have
𝑃
𝑙
𝐴
𝐵
𝐶
∠𝑃𝐴𝐶 = ∠𝑃𝐴𝐵 < ∠𝑃𝐵𝐶,
which is a contradiction.
Therefore the perpendicular to 𝑙 from 𝑃 is unique.
Example 11.3 (Greenberg 4.1b)
∎
Show that if a triangle has a right or obtuse angle, then the
other two angles are both acute.
Proof:
Let Δ𝐴𝐵𝐶 be a triangle such that ∠𝐵𝐴𝐶 is either right or obtuse.
By Axiom (B2), we let 𝐷 be a point such that 𝐶 ∗ 𝐴 ∗ 𝐷.
It follows that ∠𝐵𝐴𝐷, being supplementary to ∠𝐵𝐴𝐶, must be
either right or acute.
By the Exterior Angle Theorem, we must have ∠𝐴𝐵𝐶 < ∠𝐵𝐴𝐷
and ∠𝐴𝐶𝐵 < ∠𝐵𝐴𝐷.
Therefore ∠𝐴𝐵𝐶 and ∠𝐴𝐶𝐵 are both acute angles.
𝐵
𝐶
𝐴
𝐷
∎
AAS Theorem and RHS Theorem for congruence of triangles are consequences of the Exterior
Angle Theorem. Note that “SSA” is in general not a valid criterion for congruence of triangles,
unless when the pair of congruent angles are right or obtuse angles.
Page 3 of 11
MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Tutorial Note 7
Mar 28, 2012 (Week 9)
Although two parallel lines may not have a common perpendicular in general, it can be shown
that if two lines have congruent alternate angles over a transversal, then they have a common
perpendicular indeed. This common perpendicular may not be unique though.
Example 11.4 (Wallace & West 3.4.3) Let 𝑙 and 𝑚 be two distinct lines such that 𝐴 ∗ 𝐵 ∗ 𝐶
are incident to 𝑙, 𝐴′ ∗ 𝐵 ′ ∗ 𝐶 ′ are incident to 𝑚, all the above six points are
⃡ ′ . Show that if
distinct, and 𝐴 and 𝐴′ are on different sides of 𝐵𝐵
∠𝐴𝐵𝐵 ′ ≅ ∠𝐴′ 𝐵 ′ 𝐵, then there exists a line perpendicular to both 𝑙 and 𝑚.
Proof:
We consider the following two cases:
(i) If ∠𝐴𝐵𝐵 ′ ≅ ∠𝐴′ 𝐵 ′ 𝐵 are both right angles, then the line ⃡𝐵𝐵 ′ itself is perpendicular to
both 𝑙 and 𝑚.
(ii) If ∠𝐴𝐵𝐵 ′ ≅ ∠𝐴′ 𝐵 ′ 𝐵 are not right angles, then without loss of generality we may assume
that they are acute angles (if they were obtuse, then we consider ∠𝐶𝐵𝐵 ′ ≅ ∠𝐶 ′ 𝐵 ′ 𝐵
instead).
We let 𝑀 be the mid-point of 𝐵𝐵 ′, and let 𝑃 and 𝑃′
𝐴
𝑃 𝐵
𝐶
𝑙
be the feet of perpendicular on 𝑙 and 𝑚 from 𝑀
respectively. It clearly follows that
𝑀
 𝐵 ≠ 𝑃 and 𝐵 ′ ≠ 𝑃′ ; and
𝑚
𝐶′
𝐵 ′ 𝑃 ′ 𝐴′
 𝑃 and 𝐴 are on the same side of ⃡𝐵𝐵 ′, and 𝑃′ and
⃡ ′.
𝐴′ are on the same side of 𝐵𝐵
(Otherwise applying the Exterior Angle Theorem to Δ𝐵𝑃𝑀 or Δ𝐵 ′ 𝑃′ 𝑀′ , we have
∠𝐴𝐵𝐵 ′ = ∠𝐴𝐵𝑀 > ∠𝐵𝑃𝑀 or ∠𝐴′ 𝐵 ′ 𝐵 = ∠𝐴′ 𝐵 ′ 𝑀 > ∠𝐵 ′ 𝑃′ 𝑀, a contradiction.)
Our aim is to show that ⃡𝑴𝑷 = ⃡𝑴𝑷′ .
Since 𝐵𝑀 ≅ 𝐵 ′ 𝑀, ∠𝐵𝑃𝑀 ≅ 90° ≅ ∠𝐵 ′ 𝑃′ 𝑀 and ∠𝑃𝐵𝑀 = ∠𝐴𝐵𝐵 ′ ≅ ∠𝐴′ 𝐵 ′ 𝐵 = ∠𝑃′ 𝐵 ′ 𝑀,
it follows that Δ𝐵𝑃𝑀 ≅ Δ𝐵 ′ 𝑃′ 𝑀 by AAS Theorem.
In particular, we have ∠𝐵𝑀𝑃 ≅ ∠𝐵 ′ 𝑀𝑃′ .
Since 𝐵 ∗ 𝑀 ∗ 𝐵 ′ , it follows that ∠𝐵𝑀𝑃′ is a supplementary angle to ∠𝐵 ′ 𝑀𝑃′ . Thus
∠𝐵𝑀𝑃′ is a supplementary angle to ∠𝐵𝑀𝑃, which implies that 𝑃 ∗ 𝑀 ∗ 𝑃′ .
⃡ = ⃡𝑀𝑃′ is a line perpendicular to both 𝑙 and 𝑚.
Therefore 𝑀𝑃
∎
Page 4 of 11
MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Tutorial Note 7
Mar 28, 2012 (Week 9)
The Angle-side Correspondence Theorem states that a greater side corresponds to a greater
opposite angle, and vice versa. This holds within one single triangle, as well as when
comparing two triangles with two pairs of congruent sides.
Example 11.5 (Greenberg 4.22, 23) Suppose that 𝐴 ∗ 𝐵 ∗ 𝐶.
(a) If 𝐷 is a point not incident to ⃡𝐴𝐶 such that ⃡𝐷𝐶 ⊥ ⃡𝐴𝐶 , show that
𝐷𝐴 > 𝐷𝐵 > 𝐷𝐶.
(b) If 𝐸 is a point not incident to ⃡𝐴𝐶 , show that either 𝐸𝐵 < 𝐸𝐴 or
𝐸𝐵 < 𝐸𝐶.
Proof:
(a) Let 𝑃 be a point such that 𝐴 ∗ 𝐶 ∗ 𝑃.
Since ⃡𝐷𝐶 ⊥ ⃡𝐴𝐶 , it follows that ∠𝐷𝐶𝐵 ≅ ∠𝐷𝐶𝑃 are right angles.
𝐴 𝐵
Now since
(Exterior Angle Theorem, Δ𝐵𝐷𝐶),
∠𝐷𝐶𝐵 ≅ ∠𝐷𝐶𝑃 > ∠𝐷𝐵𝐶
𝐷
𝐶
𝑃
we have 𝐷𝐵 > 𝐷𝐶 by applying Angle-side Correspondence Theorem to Δ𝐵𝐷𝐶.
Moreover, ∠𝐷𝐵𝐶 is acute. Since ∠𝐷𝐵𝐴 is the supplementary angle of ∠𝐷𝐵𝐶, ∠𝐷𝐵𝐴
is obtuse. Now since
(Exterior Angle Theorem, Δ𝐴𝐷𝐵),
∠𝐷𝐵𝐴 > ∠𝐷𝐵𝐶 > ∠𝐷𝐴𝐵
we have 𝐷𝐴 > 𝐷𝐵 by applying Angle-side Correspondence Theorem to Δ𝐴𝐷𝐵.
∎
⃡ . Then we consider the following three
(b) Let 𝐹 be the foot of perpendicular from 𝐸 to 𝐴𝐶
cases:
⃡ . So from the result of (a) we have 𝐸𝐴 > 𝐸𝐵.
⃡ ⊥ 𝐴𝐵
(i) If 𝐹 = 𝐵, then 𝐸𝐵
(ii) If 𝐴 ∗ 𝐵 ∗ 𝐹, then ⃡𝐸𝐹 ⊥ ⃡𝐴𝐹 . So from the result of (a) we have 𝐸𝐴 > 𝐸𝐵(> 𝐸𝐹).
⃡ . So from the result of (a) we have 𝐸𝐶 > 𝐸𝐵(> 𝐸𝐹).
(iii) If 𝐹 ∗ 𝐵 ∗ 𝐶, then ⃡𝐸𝐹 ⊥ 𝐶𝐹
In any case we have either 𝐸𝐵 < 𝐸𝐴 or 𝐸𝐵 < 𝐸𝐶.
∎
Example 11.6 Let Δ𝐴𝐵𝐶 be a triangle, 𝐷 be a point such that 𝐵 ∗ 𝐷 ∗ 𝐶 and 𝐴𝐷 is the
angle bisector of ∠𝐵𝐴𝐶. Show that 𝐴𝐵 < 𝐴𝐶 if and only if 𝐷𝐵 < 𝐷𝐶.
Proof:
(⇒) Since 𝐴𝐵 < 𝐴𝐶, by Segment Relocation Axiom (C3), there exists a point 𝐸 such that
𝐴 ∗ 𝐸 ∗ 𝐶 and 𝐴𝐵 ≅ 𝐴𝐸.
Now consider the triangles Δ𝐴𝐵𝐷 and Δ𝐴𝐸𝐷 . Since 𝐴𝐵 ≅ 𝐴𝐸 , 𝐴𝐷 = 𝐴𝐷 and
∠𝐵𝐴𝐷 ≅ ∠𝐶𝐴𝐷 = ∠𝐸𝐴𝐷, we have Δ𝐴𝐵𝐷 ≅ Δ𝐴𝐸𝐷 by SAS Axiom.
In particular, we have 𝐷𝐵 ≅ 𝐷𝐸 and ∠𝐴𝐷𝐵 ≅ ∠𝐴𝐷𝐸.
Page 5 of 11
MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Tutorial Note 7
Mar 28, 2012 (Week 9)
Thus,
(Exterior Angle Theorem, Δ𝐴𝐸𝐷)
∠𝐶𝐸𝐷 > ∠𝐴𝐷𝐸
≅ ∠𝐴𝐷𝐵
(Exterior Angle Theorem, Δ𝐴𝐶𝐷)
> ∠𝐴𝐶𝐷
= ∠𝐸𝐶𝐷.
Applying the Angle-side Correspondence Theorem to Δ𝐶𝐷𝐸,
we have 𝐷𝐸 < 𝐷𝐶. Therefore 𝐷𝐵 ≅ 𝐷𝐸 < 𝐷𝐶.
𝐵
𝐴
𝐸
𝐷
𝐶
(⇐) The converse follows easily from trichotomy of congruence classes of segments:
Suppose that 𝐷𝐵 < 𝐷𝐶.
There can only be three cases: 𝐴𝐵 > 𝐴𝐶, 𝐴𝐵 ≅ 𝐴𝐶, or 𝐴𝐵 < 𝐴𝐶.
(i) If 𝐴𝐵 > 𝐴𝐶, then exchanging 𝐵 and 𝐶 in the last paragraph, we have 𝐷𝐵 > 𝐷𝐶.
(ii) If 𝐴𝐵 ≅ 𝐴𝐶 , then together with ∠𝐵𝐴𝐷 ≅ ∠𝐶𝐴𝐷 and 𝐴𝐷 = 𝐴𝐷 , we must have
Δ𝐴𝐷𝐵 ≅ Δ𝐴𝐷𝐶 by SAS Axiom. In particular, we have 𝐷𝐵 ≅ 𝐷𝐶.
In each of these two cases, there is a contradiction. So we must have 𝐴𝐵 < 𝐴𝐶.
∎
In this course, in order to make our arguments independent of “units of measurement”, we do
not explicitly define the notions of the “length” of a segment and the “size” of an angle.
However, to avoid the clumsy repetition of using the Relocation and Addition Axioms, we will
define two addition operations (+) on the set of congruence classes of segments and angles
respectively. These two addition operations are both commutative and associative.
Moreover, subtraction (−) can also be defined in the sense of the two Subtraction Theorems
(but not in the usual sense of “adding the additive inverse”!).
Remark:
(1) In order for angle addition to make sense, each of the two summands must be
smaller than or congruent to the supplementary angle of the other.
(2) (Important!) Always remember to justify that 𝑨 ∗ 𝑩 ∗ 𝑪 before you write
“𝐴𝐶 ≅ 𝐴𝐵 + 𝐵𝐶”, and remember to justify that 𝑫 is in the interior of ∠𝑩𝑨𝑪
(i.e. 𝐴𝐵 ∗ 𝐴𝐷 ∗ 𝐴𝐶 ) before you write “∠𝐵𝐴𝐶 ≅ ∠𝐵𝐴𝐷 + ∠𝐷𝐴𝐶”!
We often use the symbol “90°” to denote the congruence class represented by a right angle.
Similarly, “180°” is used as a symbol which denotes the congruence class represented by the
sum of two supplementary angles. It should be emphasized that we do not mean that there is
a unit of measurement “°” for the “size” of an angle.
Page 6 of 11
MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Tutorial Note 7
Mar 28, 2012 (Week 9)
The Triangle Inequality says that the sum of any two sides of a triangle is greater than the third.
One notable fact about this result is that it can be proved without using any continuity axiom,
but its converse is equivalent to the Circle-circle Continuity Theorem.
Example 11.7 (Prenowitz & Jordan 14.II.6)
Let 𝑃 be a point in the interior of a triangle
Δ𝐴𝐵𝐶. Show that 𝐴𝑃 + 𝑃𝐵 < 𝐴𝐶 + 𝐶𝐵 and ∠𝐴𝑃𝐵 > ∠𝐴𝐶𝐵.
Proof:
Since 𝑃 is in the interior of Δ𝐴𝐵𝐶, it is in the interior of ∠𝐵𝐴𝐶
in particular.
𝐶
𝐷
So we have 𝐴𝐵 ∗ 𝐴𝑃 ∗ 𝐴𝐶 , and by Crossbar Theorem, 𝐴𝑃 meets
𝐵𝐶 at some point 𝐷, where 𝐶 ∗ 𝐷 ∗ 𝐵.
Since 𝑃 is also in the interior of ∠𝐴𝐵𝐶 = ∠𝐴𝐵𝐷, it follows that
𝑃
𝐴
𝐵
𝐵𝐴 ∗ 𝐵𝑃 ∗ 𝐵𝐷, and so 𝐴 ∗ 𝑃 ∗ 𝐷.
Clearly 𝛥𝐴𝐶𝐷 and Δ𝐵𝐷𝑃 both exist.
Now we show that 𝐴𝐶 + 𝐶𝐵 > 𝐴𝑃 + 𝑃𝐵. This follows easily:
(𝐶 ∗ 𝐷 ∗ 𝐵)
𝐴𝐶 + 𝐶𝐵 ≅ 𝐴𝐶 + 𝐶𝐷 + 𝐷𝐵
(Triangle Inequality, Δ𝐴𝐶𝐷)
> 𝐴𝐷 + 𝐷𝐵
(𝐴 ∗ 𝑃 ∗ 𝐷)
≅ 𝐴𝑃 + 𝑃𝐷 + 𝐷𝐵
(Triangle Inequality, Δ𝐵𝐷𝑃)
> 𝐴𝑃 + 𝑃𝐵
Finally, we show that ∠𝐴𝑃𝐵 > ∠𝐴𝐶𝐵. This is also easy:
(Exterior Angle Theorem, Δ𝐵𝐷𝑃)
∠𝐴𝑃𝐵 > ∠𝑃𝐷𝐵
(𝐴 ∗ 𝑃 ∗ 𝐷)
= ∠𝐴𝐷𝐵
(Exterior Angle Theorem, Δ𝐴𝐶𝐷)
> ∠𝐴𝐶𝐷
(𝐶 ∗ 𝐷 ∗ 𝐵)
= ∠𝐴𝐶𝐵
∎
The sum of any two angles of a triangle is always less than 180°. This fact is useful in some of
the exercise problems. The sum of all three angles of a triangle is an interesting topic that we
will cover next week.
Page 7 of 11
MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Tutorial Note 7
Mar 28, 2012 (Week 9)
Exercise
1. (Noronha 1.5.2 adapted) A rhombus is a parallelogram with all sides congruent.
(a) Show that if all sides of a quadrilateral are congruent, then it is a rhombus.
(b) Show that the diagonals of a rhombus bisect opposite angles.
(c) Show that the diagonals of a rhombus bisect each other.
(d) Show that the diagonals of a rhombus are perpendicular to each other.
2. (Prenowitz & Jordan 14.II.9) Show that a point cannot be equidistant from three distinct points
incident to the same line.
3. Given a circle 𝑐, a line 𝑙 is said to be tangent to 𝑐 if it intersects 𝑐 at exactly one point, i.e.
𝑙 ∩ 𝑐 = {𝑃} for some point 𝑃. Now let 𝑐 be a circle centered at a point 𝑂, 𝐴 ∈ 𝑐, and 𝑙 is
a line incident to 𝐴. Show that 𝑙 is tangent to 𝑐 if and only if 𝑙 is perpendicular to ⃡𝐴𝑂.
4. (Greenberg 4.18) Show that the three angle bisectors of any triangle are concurrent.
Hint:
First show that two of the angle bisectors must meet at a point in the interior of the
triangle. Then show that this point also lies on the third angle bisector by showing
that it is equidistant from the sides.
Remark: This point is called the incenter of the triangle.
5. (Noronha 1.5.10) Let Δ𝐴𝐵𝐶 be a triangle such that ∠𝐵𝐴𝐶 is obtuse.
between 𝐶 and the foot of perpendicular from 𝐵 to ⃡𝐴𝐶 .
Show that 𝐴 is
6. Prove the converse to the result from Example 11.4: Let 𝑙 and 𝑚 be two distinct lines that
have a common perpendicular. Show that there exist transversals other than the common
perpendicular which cut 𝑙 and 𝑚 so as to form congruent alternate angles.
7. Let Δ𝐴𝐵𝐶 be a triangle and 𝐷 be the mid-point of the side 𝐵𝐶. Show that ∠𝐵𝐴𝐷 > ∠𝐶𝐴𝐷
if and only if 𝐴𝐵 < 𝐴𝐶.
8. (Greenberg 4.11) It has been shown in the lecture (or actually in high school) that the
Euclidean Parallel Postulate implies the converse of the Alternate Angle Theorem. Now show
that the converse of the Alternate Angle Theorem implies the Euclidean Parallel Postulate.
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