Download MAT360 Lecture 8

MAT 360 Lecture 8
Neutral Geometry
Remarks
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Midterm Grades
Next Tuesday: Sketchpad homework
11/13: Chapter 4: 1,3,4,5,6 15 (estimated)
11/20: Sketchpad project
11/27: Proof project:
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Choose one of the theorems we studied and write
a detailed proof.
The proof should have at least 10 steps.
Each step must be numbered and justified.
12/13: Chapter 5: 8 Chapter 6:: 2,3,5,14
(estimated)
Euclid’s postulates (modern formulation)
I.
II.
III.
IV.
V.
For every point P and every point Q not equal to P
there exists a unique line l that passes for P and
Q.
For every segment AB and for every segment CD
there exists a unique point E such that B is
between A and E and the segment CD is
congruent to the segment BE.
For every point O and every point A not equal to
O there exists a circle with center O and radius
OA
All right angles are congruent to each other
For every line l and for every point P that does not
lie on l there exists a unique line m through P that
is parallel to l.
In Chapter 3,
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we will study results that can be proved using
only the axioms of incidence, betweenness,
congruence and continuity.
That is, all Hilbert’s axioms with the
exception of the axiom of parallelism.
Neutral geometry consists in all the results
that can be proved in this way.
Exercise:
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Let t be a line intersecting the lines l and l’.
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Define alternate interior angles.
Theorem (Alternate Interior Angle
Theorem)
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If two distinct lines cut by a transversal have
a pair of congruent alternate interior angles
then the two lines are parallel.
Corollaries
1. Two distinct lines perpendicular to the same
line are parallel.
2. (consequence of 1.) The perpendicular
dropped from a point P not on the line l to l,
is unique.
3. If l is a line and P is a point not on l, then
there exists at least one line through P
parallel to l.
Question Can you prove…
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If two lines are parallel and they are
intersected by a third line, then the alternate
interior angles are congruent.
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Compare this statement with the Alternate
Interior Angle Theorem: If two distinct lines
cut by a transversal have a pair of congruent
alternate interior angles then the two lines
are parallel.
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Definition
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An angle supplementary to an angle of a
triangle is called an exterior angle.
The two angles no adjacent to an exterior
angle are called remote interior angles (of
that exterior angle).
Theorem: Exterior angle theorem
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An exterior angle is greater that either remote
interior angle.
Consequences
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(SAA) Given two triangles, ΔABC and ΔDEF.
If AC ~ DF, <A ~ <D and <B ~ <E then the
two triangles are congruent.
Two right triangles are congruent if the
hypotenuse and a leg of one are congruent
to the hypotenuse of a leg of the other.
Every segment has a unique midpoint.
Every angle has a unique bisector
Every segment has a unique perpendicular
bisector
Theorem (Measure of angles)
There is a unique way to assign a degree
measure to each angle with the following
properties.
1. (<A)° is a real number and 0°< (<A)° <180°
2. (<A)°= 90° if and only if <A is a right angle.
3. (<A)°= (<B)° if and only if <A ~<B
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Observe that we denote the measure of an
)<A angle by (<A)°
Theorem (Measure of angles, cont)
4. If AC is a ray interior to <DAB then
(<DAB)°= (<DAC)°+ (<CAB)°.
5. For every real number x such that 0<x<180,
there exists an angle <A such that (<A)°= x°
6. If <B is supplementary to <A then
(<A)°+(<B)°=180°
7. (<A)° < (<B)° if and only if (<A) < (<B)
Idea of the proof
1. We “know” which angles measure 90°
2. We can bisect angles, so we find angles
that measure 45° and 90 °+45°
3. By bisecting again and “adding” we find
angles of 22.5°, 45°+22.5°,
90°+22.5°,135°+22.5°,
4. And so on…
5. Using continuity one can show that it is
possible to “squeeze” all the other angles
between two of the ones above
Theorem (Measure of segments)
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1.
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Given a segment OI (called a unit segment)
there is a unique way of assigning a length
AB to each segment such that the following
property holds
OI =1
AB is a positive real number
AB=CD if and only if AB ~ CD
A*B*C if and only if AC=AB+BC
AB<CD if and only if AB<CD
For every positive real number x, there exist
a segment AB such that AB=x.
Corollary
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The sum of two angles of a triangle is less
than 180°
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Exercise: Prove it using Exterior Angle
Theorem and Measure of Angles Theorem.
Saccheri-Legendre Theorem
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The sum of the interior angles of a triangle is
less than or equal to 180°.
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Idea of the proof: Use that given a triangle
ΔABC, there exists a triangle ΔDEF, such
that <E measures half of <A and the sum of
the interior angles of ΔDEF is equal to ΔABC.
Needs Arquimedean property of real
numbers.
Corollary
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The sum of the measures of two angles of a
triangle is less than or equal to the degree
measure of their remote exterior angle.
Definition
Quadrilateral □ABCD is convex if it has a
pair of opposite sides, let’s say AB and CD
such that
1. AB is contained in one of the half planes
bounded by the line CD
2. CD is contained in one of the half planes
bounded by the line AB.
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When conditions 1 and 2 hold we say that
segments AB and CD are semiparallel.
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Corollary of Saccheri-Legendre Theorem
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The sum of the interior angles in any convex
quadrilateral is less than or equal to 360°
Proof: Step 1: Show that if □ABCD is a
convex quadrilateral then both pairs of
opposite sides are semiparallel. In other
words, AB and CD are semiparallel and BC
and AD are semiparallel.
Step 2: Find triangles to apply SaccheriLegendre.
Recall
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Euclid’s Postulate V: If
two lines are
intersected by a
transversal in such a
way that the sum of the
degree measures of
the two interior angles
on one side of the
transversal is less than
180° then the two lines
intersect on that side of
the transversal.
Hilbert’s
Axiom of
Parallelism: For every line l
and for every point P not in
l there exists at most one
line m through P such that
l is parallel to m.
Theorem
1. If Euclid’s Postulate V holds then Hilbert’s
Axiom of Parallelism holds.
2. If Hilbert’s Axiom of Parallelism then
Euclid’s Postulate V holds
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In other words
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Euclid’s Postulate V if and only if Hilbert’s
Axiom of Parallelism
All the following statements are equivalent
to Hilbert Parallel Postulate
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If line intersects one of two parallel lines
then it intersects the other.
If two lines are parallel then alt. int. angles
are congruent
If t is a transversal to a line l and m, l is
parallel to m and t is perpendicular to l then t
is perpendicular to m.
If k and l are parallel lines, m is
perpendicular to k and n is perpendicular to l
the m=n or m is parallel to n.
Definition
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The angle sum of the triangle ΔABC is
(<A)°+ (<B)°+ (<C)°
The defect of ΔABC, denoted by δABC is
180°-( (<A)°+ (<B)°+ (<C)°)
Question: Which values can be taken by the
defect of a triangle?
Theorem:
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Let ΔABC be a triangle and let D be a point
between A and B. Then
δABC = δACD + δBCD
Corollary
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Let ΔABC be a triangle and let D be a point
between A and B. Then
δABC = 0 if and only if
δACD=0 and δBCD=0.
Equivalent definitions of rectangle in
Euclidean Geometry
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A quadrilateral with four right angles
A quadrilateral with four angles congruent to
each other
A parallelogram with at least one right angle.
Definition
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A rectangle is a quadrilateral whose four
angles are right angles.
Theorem
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If a triangle with an angle sum of 180° exists
then a rectangle exists.
If a rectangle exists then every triangle has
angle sum of 180°.
The steps of the proof
1. Construct a right triangle with defect 0.
2. Construct a rectangle
3. Given a right triangle ΔABC where <ABC is
the right angle, construct a rectangle
□DEFG such that AB < DE and BC<EF.
4. Prove that all right triangles have defect 0.
5. Prove that all triangles have defect 0.