Download Geometry 1st Grading Period Notes 083111 Pointers Topics History

Geometry
1st Grading Period Notes
083111
Pointers
Topics
History to Proving excluding secondary triangle parts
Type of Tests
1. True or False (10)
2. Always, sometimes, never (10)
3. Matching type (15)
4. Alternate choice (10)
5. Illustration (10)
6. Puzzle (10)
7. Problem Solving (20)
8. Proving (15)
I. Geometry and History
Geometry
 Greek words “ge” (geos, geo, gaia) meaning earth
and “metron” (metrein, metre, metria)meaning
measure
 branch of mathematics dealing with properties
 measurements, and relations of points, lines, angles,
planes, and solids
 points, lines, planes, solids
Prehistoric
 informal Geometry of shape
 early stone age
 geometric figures such as circles, squares, and
triangles can be found in wall paintings
Ancient Egypt and Babylonia
 intuitive or practical geometry of size
 King Ramses Ramses II divided the Nile among his
people in return for taxes
 land surveying
 pyramid building
 because of the yearly overflowing of the Nile, the
people of Ten had to re-measure their land
Ancient Greece
 Deductive Geometry
 Practical aspect of Geometry
 More on reasoning
 Six Giants
1. Thales of Miletus
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


First Father of one of two fathers of Geometry
First among 7 wise men of Greece
First to prove that a circle is bisected by its
diameter

Measured the height of a pyramid using the
idea on similarity and predicted an eclipse of
a sun

Founded the Ionian school
2. Pythagoras of Samos

Founded the “Pythagoras”

Discovered the sum of interior angles is a
triangle is 180

Pythagorean Theorem
3. Plato

Founded “the academy” in 387 BC

Developed a theory of forms, in his book
“Phaedo” which considers mathematical
objects as perfect forms
4. Aristotle

Most famous student of Plato

Tutors of Alexander the Great

Noted the difference between an axiom and a
postulate

Developed the laws of logical reasoning
5. Euclid

Transformed geometry from the science of
measurement to science of reasoning

Second father of geometry

Best known for his 13 books treatise

“the Elements” collecting the theorems of
Pythagoras, Hippocrates, Theatetus, Euxodus,
and other predecessors into a logically
connected whole
6. Archimedes

Greatest of Greek mathematicians and was
also an inventor of many mechanical devices
(including the screw, the pulley, and the
lever)- physics

Pi, circumference

Proved that the volume of a cube is 2/3 the
volume of a circumscribed cylinder
7. Hippocrates

Wrote the first “elements of Geometry” which
Euclid may have used as a model

First to show that the ratio of the squares is of
their radii
Geometry in Medieval and Modern times
 Apollonius
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
Introduced the terms parabola, ellipses, and
hyperbola in his famous book the “conics”

The great geometer
 Rene Descartes

Coordinate system or coordinate on analytical
geometry- Cartesian plane

First to use letters to represent numbers
 Blaise Pascal

Published an essay on conic sections

Pascals triangle
II. Reasoning
 Involves the derivation of conclusion that is
supported by a statement
 Kinds
1. Intuition- Guessing, does not require
mathematical proofs, knowing without the
use of reasoning or rational process
2. Analogy- alike in one aspect, will always be
alike in all aspects, reasoning by
comparison
3. Induction- specific to general, allows to
reach conclusions based on a pattern of
specific events
4. Deduction- general to specific, also the
process of drawing a conclusion form
something known or assumed
III. Conditional Statements
Conditionals
 statements written in if-then from
 if gives the hypothesis and then gives the conclusion
Converse of a Conditional
 formed by interchanging the hypothesis and the
conclusion
Inverse of a Conditional
 formed by negating both the hypothesis and the
conclusion
 formed by using the word not, changing = to ≠
Contrapositive of a Conditional
 formed by negating the hypothesis and the
conclusion of its converse
III. Terms in Geometry
Undefined Terms

no formal definition

can only be represented or described

can also be dealt with through their properties
which are taken up as postulates
 Point

represented by a dot

named by a capital letter
 points contained on the same line

Non collinear Points
 points not contained on the same line

Coplanar Points
 points contained on the same plane

Non coplanar Points
 points not contained on the same plane
 Line

set of points extending infinitely in both directions

has length but no width nor thickness

arrowhead

named using 2 points on each end or a small letter
in the middle

Line Segment
 set of points in between two points
 named by its endpoints
 definite

Ray
 part of a line with one endpoint and extends
infinitely in the other direction
 named by two points, name starting at end points
 Opposite Rays
 two rays with a common endpoint contained on
the same line but going in opposite directions
 Half Line
 set of points which is the union of all the points of
a line on one side of a given point excluding the
given point
 Plane

flat surface

usually represented by four sided figures

has infinite length and width but no thickness

named by 3 non collinear points or a capital letter
on the corner
Point
Line
Line Segment
Ray
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Opposite Rays
Half Lines
Plane
A
Basic Terms
 Postulates

assumed to be true

also known as an Axiom

accepted without proof
 Theorems

further need to be proved
 Space

set of all points
 Collinear

points are located on the same line
 Non Collinear

points are not located on the same line
 Coplanar

points are located on the same plane
 Non Coplanar

points are not located on the same plane
 Congruent Segments

segments with the same length
 Midpoint of a Segment

divides the segment into two congruent parts

also the bisector of a segment
 Betweenness

not all betweenness are midpoints but all
midpoints are betweenness
 Bisector

midpoint of a segment and may be a line, half line,
ray, or segment passing through the midpoint

all midpoints are bisectors but not all bisectors
are midpoints
Problem
If DT= 60, find
the value of x, DS
and ST
C is the midpoint
of Segment AB.
Find AC, CB, and
AB
If RS= 3x+1, ST=
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Figure
D
S
T
A
C
B
Final Answer
x= 16
DS= 24
ST= 36
AB= 22
AC and CB= 11
x= 13
2x=2, and RT=64,
find the values of
x, RS, and ST
If RS= 87+ 4, ST=
4y+8 and RT=
157-9, find the
value of y, RS, ST
and RT
If PT= 5x+3 and
TQ= 7x=9, find x,
PT, and TQ
R
RS= 40
ST= 24
T
S
Same as figure 3
P
T
y= 7
RS= 60
ST= 36
RT= 96
x= 6
PT and TQ= 33
Q
Applying terms of if and then
F
E
C
A B
D
C
G
Hypothesis
C-D-E
BC= CD
C is the midpoint
of segment BD
Segment AC is
congruent to
segment CF
Ray FG bisects
segment BC
Conclusion
CD+ DE= CF
Segment BC is
congruent to
segment CD
Segment BC is
congruent to
segment CD
AB= CF
Reason
Betweenness
Congruence
C is the midpoint
of segment BD
Bisector
Midpoint
Congruent
Segment
W
N
X
Y
Z
Hypothesis
W-A-Z
Segment SA is
congruent
to
segment AY
A is the midpoint
of segment WZ
Ray WZ bisects
C
Conclusion
WA+ AZ= WZ
XA= AY
Reason
Betweenness
Congruent
Segments
Segment WA =
segment AZ
Segment XY is a
Betweenness
Bisector
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Segment XY
bisector of
segment WZ
Or
A is the midpoint
of segment XY
IV. Postulates and Theorems
 Midpoint Postulate

1 midpoint

Midpoint of a segment is also its bisector

Any line, segment, ray, half line of a plane is also
its bisector
 Points Postulate

A line has infinite number of points

A plane has at least 3 points

A space has 4 non coplanar points
 Line Postulate

Two points are contained in one and only one line

Through any 2 points, there exists exactly 1 line

A line consists of at least 2 points
 Plane Postulate

3 points that lie in at least one plane make various
planes

3 non collinear points that lie in exactly 1 plane
 Space Postulate

Space is determined by 4 non collinear points
 Flat Plane Postulate

If 2 points on a line lie on a plane, then the line lies
on the same plane
 Plane Intersection Postulate

Forms a line with a currently unknown variable as
a name

If 2 planes intersect then their intersection is a
line
 Intersecting Line Plane Theorem

In 2 intersecting lines, there is exactly one plane
containing both
 Point Line Plane Theorem

Given a line and a point not on the line, there is
exactly one plane containing both
 Line Intersection Theorem
 Line Plane Theorem

If a line intersects a plane that doesn’t contain if
the intersection contains only 1 point
 Distance Postulate

Distance between two points, positive number
a. Uniqueness property- there is a unique distance
between two lines
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b. Distance formula- there is a formula to solve for a
horizontal line (x-y)
c. Additive property- AB+BC=AC
 Ruler Placement Postulate

Given two points X and Y of a line, the coordinate
system can be chosen in such a way that the
coordinate of X is 0 and the coordinate of Y is
positive
 Segment construction postulate

Let ray AB be a ray, and let X be a positive number.
Then there is exactly one point P of Ray AB such
that AP=x
 Ruler postulate

To every point of the line there corresponds
exactly one real number
IV. Angles
Angle
 union of two non collinear rays with a common
endpoint
 Angle Bisector- a ray bisects an angle if it divides the
angle into two congruent angles
 Congruent Angles- angles whose measurements are
equal
 Perpendicular lines- forms 4 90 degree angles
 Perpendicular bisector- cuts into 2 congruent 90
degree parts
 Angle bisector postulate- one angle bisector in every
angle
 Angle addition postulate- add two angles to make a
bigger angle
Types of Angles
 Acute Angle : measures less than 90
 Right Angle: measure is equal to 90
 Obtuse: measure is more than 90 but less than 180
Angle Pairs
 Complementary Angles

two angles whose measures add up to 90
 Supplementary Angles

two angles whose measures add up to 180
 Adjacent Angles

two angles with a common side but do not have
common interior points
 Vertical Angles

two nonadjacent angles formed by two
intersecting lines
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 Linear Pair

two adjacent angles whose non common sides also
form a pair of opposite rays
perpendicular to
line FD
Ray FZ bisects
angle XZF
Examples
m angle FZN= m
angle XZN
A
D
B
C
Problem
If m angle ABD= 50 and m
angle CBD= 20, find m angle
ABC
If m angle ABC= 85 and m
angle ABD= 37, find m angle
CBD
If m angle ABC= 85, m angle
ABD= x+30 and m angle
CBD= 2x+10, find m angle
CBD and m angle ABD
If m angle ABD= x+25, m
angle CBD= 3x-10 and m
angle ABC- 75, what is m
angle ABD and m angle CBD
If m angle ABC- 4x+ 40, m
angle ABD= 50, and m angle
CBD= 2x+ 10, find m angle
CBD and m angle ABC
The m angle ABD= x+5, m
angle CBD= 2x-20 and m
angle ABC= 45. Find the m
angle ABD and m angle CBD
The m angle CBD is less that
twice the m angle ABD. If m
angle ABC is 80, what is the
m angle CBD
The measure of angle ABC
and angle CBD are in ratios
of 2:3. If the m angle ABC is
70, what is the measure of
angle ABD? m angle CBD
Final Answer
Angle ABC= 70 degrees
Angle CBD= 48 degrees
Angle FZN is
congruent to
angle XZN
Angle FZN is
congruent to
angle XZN
m angle XZG= m
angle FZY
Perpendicular
bisector
Ray FZ bisects
angle XZF
Angle bisector
Congruent angles
Congruent angles
Angle XZF, angle
XZG, angle GZY
and angle YZF are
right angles
Perpendicularity
m angle CBD= 40 degrees
m angle ABD= 45 degrees
Type of Angle
Complimentary Angles
m angle ABD= 40 degrees
m angle CBD= 35 degrees
Supplementary Angles
Figure
Adjacent Angle
m angle CBD= 30 degrees
m angle ABC= 80 degrees
Vertical Angles
m angle ABD= 25 degrees
m angle CBD= 20 degrees
Vertical Angles
m angle CBD= 50 degrees
Linear Pair
m angle ABD= 28 degrees
m angle CBD= 42 degress
V. Convex Sets
Applying terms on angles
Hypothesis
Line XY is
Angle XZG is
congruent to
angle FZY
Line FG is the
perpendicular
bisector of line
XY
Line XY is
perpendicular to
line FG
angles
Conlcusion
Forms four right
Reason
Perpendicularity
Convex Set
 a given set of points is convex if there is a segment
determined by any two points of the set such that
this segment lies entirely in the given set
Concave Set
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 a given set of points is concave or non convex if
there is a segment determined by any two points of
the set such that this segment does lies entirely in
the given set
1.
2.
3.
Line Separation Postulate- a point separates a line into
two half- lines, each of which is a convex Set
Plane Separation Postulate- line of separation or edge
divides into two half planes
Space separation postulate- a plane separates into 2
half spaces, each of which is a convex set
Convex
Concave
B
A
A
B
VI. Triangles
Triangle
 union of three segments determined by three non
collinear points
Types of Triangles
 According to Angle Measurement

Right Triangle: triangle with one right angle

Acute Triangle: triangle with 3 acute angles

Obtuse Triangle: triangle with one obtuse angle

Equiangular Triangle: triangle with three
congruent angles
 According to Measure of its Sides

Scalene Triangle: triangle wherein no two sides
are congruent

Isosceles Triangle: triangle wherein at least two
sides are congruent

Equilateral Triangle: triangle where in all three
sides are congruent
Secondary Parts of a Triangle
 Median

segment whose endpoints are a vertex of the
triangle and the midpoint of the opposite side

point of concurrency: centroid
 Altitude
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





perpendicular segment from a vertex of the
triangle to the opposite side or to the line
containing the opposite side
point of concurrency: orthocenter
Angle Bisector
point of concurrency: incenter
Perpendicular bisector
point of concurrency: circumcenter
VII. Properties





















RPE (Reflexive)
m angle 1= m angle 1
3=3
SyPE (Symmetric)
m angle 1= m angle 2
m angle 2= m angle 1
TPE (Transitive)
AB= CD, CD= EF
AB= EF
APE (Addition)
m angle 1= m angle 2; m angle 3= m angle 4
m angle 1 + m angle 3= m angle 2 + m angle 4
SPE (Subtraction)
m angle 1= m angle 2; m angle 3= m angle 4
m angle 1= m angle 3= m angle 2= m angle 4
MPE (Multiplication)
AB= CD; EF= GI
AB (EF)= CD (GI)
Substitution
A+ B= C; C=2
A+ B= 2
VIII. Writing Proofs
Proof
 set of properly ordered and accurately justified
conditional statements
 composed of a figure, the given statement, the
statement to be proven and the two-column proof
Simple Proofs
Parts of a proof
1. Given statement
2. Prove statement
3. T- Column
4. Figure
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Examples
Angle RCB is congruent to
angle BCV
Ray CB bisects angle RCV
 G: B is the midpoint of segment AC
 P: Segment AC is congruent to Segment BC
A
B
 G: AB= CD
 P: AC= BD
C
Statement
B is the midpoint of
segment AC
Segment AC is congruent to
segment BC
AB= BC
Definition of midpoint
Congruent Segments
R
 G: Ray CB bisects angle RCV
C
 P: m angle ACB = m angle BCV
V
Reason
Given
Angle bisector
Congruent angles
C
Statement
Line CD bisects segment AB
E is the midpoint of
segment AB
Segment AE is congruent to
segment EB
AE= BE
Reason
Given
Bisector
Statement
m angle RCB = m angle BCV
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E
Congruent segments
Reason
Given
D
Reason
Given
RPE
APE
Bet.
Substitution
B
Statement
m angle ABD= m angle CBE
m angle DBE= m angle DBE
m angle ABD+ m angle
DBE= m angle CBE+ m
angle DBE
m angle ABE= m angle
ABD+ m angle DBE; m angle
DBC= m angle DBE+ m
angle EBC
M angle ABD= m angle CBD
A
D
E
C
Reason
Given
RPE
APE
AAP
Substitution
A
 G: AB= AE; BC= ED
 P: AC= AD
B
C
D
Midpoint
 G: m angle RBC = m angle BCV
 P: Ray CB bisects angle RCV
 Refer to example 3
C
 P: m angle ABE= m angle CBD
 Adding segments- betweenness
 Adding angles- angle addition postulate
Midpoint
A
B
B
Reason
Given
Congruent segments
 G: Line CD bisects Segment AB
 P: AE= BE
A
 G: m angle ABD= m angle CBE
R
Statement
Ray CB bisects angle RCV
Angle RCB is congruent to
angle BCV
m angle RBC is congruent to
m angle BCV
Angle bisector
Statement
AB= CD
BC= BC
AB+BC= CD+BC
AC= AB+ BC; BD= BC+ CD
AC= BD
Reason
Given
 G: Segment AB is congruent to segment BC
 P: B is the midpoint of segment AC
Statement
AB= BC
Segment AB is congruent to
Segment BC
B is the midpoint of
segment AC
Congruent angles
Statement
AB= AE; BC= ED
AB+ BC= AE+ ED
Reason
Given
APE
AC= AB+ BC; AD= AE+ ED
AC= AD
Bet.
Substitution
E
D
 G: angle GIL is congruent to angle FIJ
 P: angle GIJ is congruent to angle LIF
G
L
I
J
F
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Statement
angle GIL is congruent to
angle FIJ
m angle GIL= m angle FIJ
Reason
Given
m angle LIJ= m angle LIJ
m angle GIL + m angle LIJ=
m angle FIJ+ m angle LIJ
m angle GIJ= m angle GIL+
m angle LIJ; m angle LIF= m
angle FIJ= M angle LIJ
m angle GIJ= m angle LIF
angle GIJ is congruent to
angle LIF
RPE
APE
Congruent angles
AAP
Substitution
Congruent angles
Statement
Segment BD is congruent to
segment CE
BD= CE
Reason
Given
BD= BC+ CD; CE= CD+ DE
BC+ CD= CD+ DE
CD= CD
BC= DE
Segment BC is congruent to
segment DE
Bet.
Substitution
RPE
SPE
Congruent angles
 G: angle GHJ is congruent to angle IHK
 P: angle GHK is congruent to angle IHJ
 G: angle 1 is congruent to angle 3; angle 2 is
congruent to angle 4
 P: angle DOM is congruent to angle HAP
Statement
angle 1 is congruent to
angle 3; angle 2 is
congruent to angle 4
m angle 1= m angle 3; m
angle 2= m angle 4
m angle 1+ m angle 2= m
angle 3+ m angle 4
m angle DOM= m angle 1+
m angle 2; m angle HAP= m
angle 3+ m angle 4
m angle DOM= m angle HAP
angle DOM is congruent to
angle HAP
Reason
Given
Congruent angles
APE
AAP
Substitution
Congruent angles
 G: PR= NT
 P: PN= RT
G
K
H
Statement
angle GHJ is congruent to
angle IHK
M angle GHJ= m angle IHK
M angle GHJ= m angle GHK+
m angle KHJ; m angle IHK=
m angle KHJ+ m angle JHI
M angle GHK= m angle KHJ=
m angle KHJ+ m angle JHI
M angle KHJ= m angle KHJ
M angle GHK= m angle IHJ
Angle GHK= angle IHJ
J
I
O
Reason
Given
Congruent angles
AAP
Substitution
RPE
SPE
Congruent angles
 G: angle Tan and angle NAX are complimentary
 P: ray AX is perpendicular to ray AF
Statement
PR= NT
NR= NR
Reason
Given
RPE
PR- NR= NT- NR
PN= PR= NR; RT= NT= NR
PN= RT
SPE
Bet.
Substitution
 G: segment DB is congruent to segment CE
 P: segment BD is congruent to segment DE
A
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Congruent segments
B
C
D
E
T
N
A
Statement
Angle Tan and angle NAX
are complimentary
M angle TAN= m angle
NAX= 90
M angle TAX= m angle
TAN= m angle NAX
M angle TAX= 90
Angle TAX is a right angle
Ray AX is perpendicular to
ray AT
X
Reason
Given
Complimentary angles
AAP
Substitution
Right angle
Perpendicular Lines
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