Download I. Chapter One: Points, Lines, and Planes

I. Chapter One: Points, Lines, and Planes
 Definitions
a. The three undefined terms are points, lines and planes
 Points: Ex.
H
 Lines: AB, AB, AB, AB stands for Line AB, Ray AB, Segment AB and the
length of Segment AB
b. Two points define a line
c. Three non-collinear points define a plane
 Intersections:
a. When two lines intersect they form a point
b. When two planes intersect they form a line
 Collinear:
A point is collinear with a line if it lies on that line. For example we can show that
(3, -5) is collinear with the line y  3 x  14 by substituting in. -5 = 3 (3) – 14
-5 = 9 – 14
-5 = -5
 Distance
a. Number Line: d  a  b , where a and b are the coordinates of the two points
b. Coordinate Plane:
two points
d  ( x2  x1 ) 2  ( y 2  y1 ) 2 , x and y are the coordinates of the
c. Three-Dimensional: d  ( x2  x1 ) 2  ( y 2  y1 ) 2  z 2  z1 
B
Midpoint
A
a. If B is the midpoint of AC then AB = BC
ab
b. Number Line: midpt 
2
 x  x2 y1  y 2 
c. Coordinate Plane: xmidpt , y midpt    1
,

2 
 2
d. This is like finding the average of the two endpoints.
 x  x2 y1  y 2 z1  z 2 
e. Three-Dimensional: xmidpt , y midpt , z midpt    1
,
,

2
2 
 2
2

C
Exercises:
1. R, S, and T are collinear. S is between R and T. RS  2w  1 , ST  w  1 , and RT  18 . Solve for w, and
then determine the length of RS .
2. Find the distance between the points (-1, 2) and (-3, 4).
3. The intersection of two planes is a(n) ______.
4. L is the midpoint of JQ. If J(2, -4) and Q(-6, -8), find L.
5. Which of the following points is collinear with the line y  2 x  3 ?
A. (2, 3)
B. (-11, 22)
C. (-3, -3)
D. (1, -5)
A
N
T
J
G
S
6. What is the intersection of planes TAN and UTJ.
U
M
7. If E is between H and L, find x and HE.
EH = 6x – 5, EL = 2x + 3, HL = 9x – 6
8. If mABC  5x and mCBD  11x  12. If BC is an angle bisector find the mABC.
C
A
D
B
II. Chapter Two: Deductive Reasoning and Introduction to Proofs

Vertical Angles
Vocabulary
a. Vertical Angles: angles “across from each other”, formed by two intersecting lines
Relationship: Congruent 1  2 because they are vertical angles
2
1
Linear Pair
Angle Bisector
F
1
b. Linear Pair: angles that “form a line” and are supplementary
Relationship: Supplementary; m1  m2  180
2
E
G
c. Angle Bisector: line/segment/ray that “cuts an angle in half, left side = right side”;
mFEG  mGED
D
Complementary Angles
d. Segment Bisector: line/segment/ray that “cuts a segment in half,
going through the midpoint, left side = right side”; HJ = JK
I
J
H
1
K
Segment Bisector
e. Supplementary Angles: two angles that add up to 180.
2
f. Complementary Angles: two angles that add up to 90; used a lot in conjunction with
perpendicular lines.
m1  m2  90
g. Straight Angle: a line, has 180 degrees
h. Right Angle: formed by two perpendicular  lines; has 90 degrees
Segment Addition Postulate
i. Segment Addition Postulate: when you add the parts of the segment together, you get
C
B
the whole thing.
A
AB + BC = AC
Angle Addition Postulate
D
j. Angle Addition Postulate: when you add the parts of an angle
together, you get the whole thing
F
E
G
k. Algebraic Properties
 Multiplication, Division, Addition, Subtraction – must be done on both sides of
the equation
 Reflective: something equals itself: a = a
 Symmetric: you can switch the equation around: 5 = x then x = 5
 Commutative: a + b = b + a
 Associative: (a + b) + c = a + (b + c)
 Distribution: a(b + c) = ab + ac
 Substitution/Transitive: when you “plug in”

Theorems vs. Postulates:
a. Postulates are statements that are accepted as true
b. Theorems are statements that must be proven to be true

Conditional Statements
a. Two parts: hypothesis and conclusion
 The hypothesis is the “if-” and is the given in a proof
 The conclusion is the “then-” and is the prove in a proof
b. Conditional:
If p then q
c. Inverse:
If not p then not q (Think prefix “in-“ means “not”)
d. Converse:
If q then p
e. Contrapositive:
If not q then not p

Algebra Proofs: Always start with the given and then solve like an algebra problem giving the
property that allows each step to happen.
Ex. 3: If
x 1
 4  4 x then x  1
2
x 1
 4  4x
2
2) x  1  2(4  4 x)
3) x  1  8  8x
4) x  9  8x
5) 9x  9
6) x  1
1)

C
A
2) Multiplication Property
3) Distributive Property
4) Addition Property
5) Addition Property
6) Division Property
Proofs
a. The Given: usually the given goes first.
b. The Picture
 If there isn’t a picture, draw one
 Write equations/relationships from the picture
 Combine these with the given statement to get step #2
D
B
1) Given
Exercises:
1
1. If m1  43, find m2.
2
5

3 x  
3
 7 then x = 18”
2. What is the hypothesis of the following conditional statement: “If 
7
B
3. Which of the following is a pair of supplementary angles?
A. BFC , GFE
B. GFC, CFD
C. EFD, FBD
D. AFB, EFA
C
A
D
E
B
6. Prove that if G is the midpoint of FH then FG + HJ = GJ
F
G
H
J
D
F
A
E
G
3( x  5)
 2 x then x  3 .
4. Prove that if
4
5. Prove that if ABC  DBE then ABD  CBE
C
n
III. Chapter 3: Parallel and Perpendicular Lines and Slope



2
Vocabulary
a. Transversal (line n in picture): a line that crosses two other lines 1 4
to form eight angles
3
p
b. Corresponding Angles ( 1 and 5; 2 and 6;
6
5
3 and 7; 4 and 8): angles that are in the same
8
7
place on the other line.
q
c. Consecutive Interior Angles (3 and 5; 4 and 6):
angles on the inside on the same side.
d. Alternate Interior Angles (3 and 6; 4 and 5)
e. Alternate Exterior Angles (2 and 7; 1 and 8
f. Slope
g. Skew lines: lines that do not intersect and are not parallel, noncoplanar lines
Slope
y  y1
a. m  2
x2  x1
b. Parallel lines have the same slope
c. Perpendicular lines have the opposite reciprocal (flip it and change the sign) slope
Proving lines parallel
a. Corresponding angles are congruent
b. Alternate interior angles are congruent (look for the Z!)
c. Alternate exterior angles are congruent
d. Consecutive interior angles (same-side interior angles) are supplementary
Exercises:
1. On an elevation chart, the ski lodge at Ski Liberty is at (4, 450) while the top of the ski lift is at (360, 1930).
What is the slope of the mountain?
2. Write the equation of a line perpendicular to 3 x  4 y  11  2 that goes through (-5, 9)?
C
F
B
3. Three segments that are skew to EF are:
G
D
H
(2x-3)
4. Find x so that the lines are parallel.
p
133
5. Given: DE II BC
3  4
Prove: 1  2
A
q
A
D
B
1
3
4
E
2
C
E
IV. Chapter 4: Proving Triangles Congruent
 Vocabulary
a. Six types of triangles:
 Obtuse: one angle has a measure > 90o
 Right: one angle measures 90o
 Side opposite right angle called the hypotenuse
 Other two sides called legs
 Acute: all angles < 90o
 Scalene: all sides have different measures
 Isosceles: at least two sides have the same measure
 Equal sides are called legs
 Third side called the base
 Angles opposite the legs are congruent and are called base angles
 Angle opposite the base is called the vertex
 Equilateral: all sides are congruent
 Also called equiangular because all angles are equal (each 60o)
 The “regular” triangle
b. Triangle Sum Thm: the interior angles in a triangle sum to 180o
c. Exterior Angle Thm: an exterior angle is equal to the sum of the two remote interior
2 1
angles. In the figure to the right, m1  m3  m4
3
d.
Ways to prove triangles congruent
4
 SSS Postulate: if the corresponding sides are  then the triangles are 
 SAS Postulate: if two sets of corresponding sides and the included angle are 
then the triangles are 
 ASA Postulate: if two sets of corresponding angles and the included angle are
 then the triangles are 
 AAS Theorem: if two sets of corresponding angles and a non-included side are
 then the triangles are 
 HL Postulate: If one set of corresponding legs and the hypotenuses are  then
the triangles are  . Note: must be right triangles
L
M
e. CPCTC: Corresponding Parts of Congruent Triangles are Congruent
 When proving triangles congruent consider these strategies:
a. Bow-tie problems: if the picture resembles a bowtie, you automatically K
S
have one set of angles congruent using vertical angles. Ex: LMK  JMN
b. Reflexive sides: when the triangles share a side, you automatically have a side using
P
reflexive. Ex: NP  NP by reflexive property.
Q
c. Parallel Lines: whenever lines are parallel, look for the Z for alt. int. angles or look
for corresponding angles.
Ex1: if RS QT then SRQ  TQU
Ex2: UV TW then VUW  TWU
N
b/c if lines are parallel then corres. s 
S
b/c if lines are parallel then alt. int. s 
T
V
U
J
Q
corresponding angles
L
M
K
R
N
U
T
W
alternate interior angles
d. Midpoint: if there is a midpoint, you know that two sides are congruent. Ex: if M is
the midpoint of LN then LM  MN
e. CPCTC: you know when to use CPCTC when what you are trying to prove is
something other than a triangle congruent to a triangle. Ex: exercise 7 below.
J
N
Exercises:
1. Classify TAY with vertices T  4,5 , A2,3 and Y 0,3 .
B
2x+1
2. Find the mACB .
A
 x+16 
x
4x - 32
D
C
3. Find the measure of x and the measure of all the angles in the triangle.
x
A
 x+29 
C
4. Find the measure of x and the measure of all angles in the triangle
B
x
(4x-63)
5. A side in a triangle is two less than 3 times the smallest side. If the third side is 6 more than twice the
smallest side and the perimeter is 34 cm., what is the measure of the longest side?
6. Given: RP CB , A is the midpoint of RB
R
P
Prove: PAR  CAB
A
B
C
7. Prove if ABD is isosceles with vertex B and BDC , BAC are right angles then BCD  BCA .
B
D
A
C
8. Prove if DG bisects EGF and EG  FG then EGD  FGD
E
D
F
9. Given: AH JK and AJ HK
H
A
Prove: AJK  KHA
J
K
G
V. Chapter 5: More on Triangles
 Vocabulary
NM is an altitude
a. Median: starts at a vertex and goes to the midpoint of the opposite side
b. Altitude: starts at a vertex and hits the opposite side forming a rt. angle
In a right triangle the legs are the altitudes
QP is a perpendicular bisector 

In an obtuse triangle you may have to extend the opposite side
P
c. Angle bisector: starts at a vertex cutting the angle in half
Q
N
M
KL is a median
K
L
QL is an angle bisector
L
d. Perpendicular bisector: starts at a midpoint and forms a right angle Q
 Triangle Inequality Theorem
a. The sum of the two shortest sides in a triangle must be greater than the third side
Ex: 3, 11, 20 do not make a 
Ex: 2, 17, 17 do make a 
Ex: 5, 16, 21 do not make a 
b. To find the range of a possible third side, you add and subtract the numbers
Ex: if two sides are 52 and 37 then the third side, x, is somewhere in the range 15 < x < 89
 Indirect Proofs
a. Take what you are trying to prove and assume the opposite
b. Work until you hit a contradiction
c. When you hit a contradiction this tells you that your assumption is false
A
d. If your assumption is false then what you are trying to prove must be true.
Ex: Given: WA  AL , mW  mL
Prove: AT is not an angle bisector of WAL
W
T
L
Proof: Assume that AT is an angle bisector of WAL . By definition, WAT  LAT . This, along with our
given ( WA  AL ) and AT  AT by reflexive leads to WAT  LAT by SAS. CPCTC then tells us that
W  L . Of course, if the angles are congruent then their measures are equal. However, this contradicts with the
given. Therefore our assumption must be false and AT is not an angle bisector.

S
17
19
T
R
18
Bigger Angle is opposite a Bigger Side
a. In a single triangle you can rank the sides if the angles are known and vice versa
S
Ex:
Since 19 > 18 > 17 then
Ex: First, subtract from 180 to get
42
mR  71 . Then since 71 > 67 > 42
mT  mS  mR
you have ST > RS > RT
T
67
R
b. In working with composite shapes (more than one triangle), only do one triangle at a
A
time and then work with the shared side
50
B
C
D
Ex: find the remaining angles:
Ex: Start with top 
71
110
65
71 > 59 > 50, meaning that
mC  38  , mADB  36 
Start with the left  : 110 > 38 > 32,
AD > AB > BD. Next the bottom
32
114
 : 65 > 60 > 55, meaning that
meaning that CB > BD > CD.
59
30
B
55
A DC > BD > BC. We can conclude
Next the right  : 114 > 36 > 30, meaning
D
that AD > AB > BD. We can conclude that CD
BD is the shortest in the figure because it is the
is the shortest in the figure because it is the only segment only segment shorter than BD, but we cant
shorter than BD, but we can’t conclude which is longest
conclude which is longest.
 Hinge Theorems
a. SAS≠: if two sets of corresponding sides are congruent, but the included angles are not
C
equal then the opposite sides are not congruent
b. SSS≠: if two sets of corresponding sides are congruent, but the third sides are not then
13
D
B
the opposite angles are not congruent
Ex: DB > CD by SAS≠
Ex: mABE  mCBD
22
A
23
by SSS≠
13
B
15
15
12
A 4 E
12
D 6
C
60
C
Exercises:
1.) If AD is a median, find the measure of AB if CD  x  4, DB  5 x  8, AB  11  x
A
2.) Using A (-3, 5), B (2, -1) and C (3, 4), write the equation of the median from A
Q
3.) Using the same triangle, write the equation of the altitude from B
R
4.) If RT is a perpendicular bisector, find the measure of QS
mQRT  3x  15
x
RS   4
5
S
P
T
5.) Determine if the following sets of sides could make a triangle:
a. 3.1, 6, 4.8
b. 11.7, 12.3, 24
c. 15, 15, 0.2
6.)
If the sides of a triangle are 22, 47, and x, describe x.
7.) What would be the first step of an indirect proof of the following statement:
“Andrew is not the shortest man in the world”
8.) Use an indirect proof for the following:
Given: AH  JK , H is not congruent J
Prove: AH is not parallel to JK
H
A
J
K
J
9.) Use an indirect proof for the following:
Given: M is the midpoint of LN , LMK is not congruent to NMJ
L
M
Prove: M is not the midpoint of KJ
N
K
V
10.) Rank the sides in order from lease to greatest
79
41
U
W
11.) Rank the angles in order from least to greatest in  PQR with P(-4, 3), Q(3, -6), and R(7, 1)
A
12.) Find the longest side:
B
10
9
8
C
D
O
13.) Prove the following:
B
Given: D is the midpoint of AC
BC  AB
Prove: m1  m2
2
A
1
D
C
B
D
C
VI. Chapter 8: Similarity
 Vocabulary
a. Similarity (~): Polygons are similar if
 their corresponding sides are proportional
 their corresponding angles are congruent
b. Scale factor
c. Proportion
 Geometric Mean
a. Algebraically: multiply and then take the square root
b. As a proportion: x  x
Ex: the geo mean b/t 6 and 10, 6  10  60, 60  2 15  7.75 or 6x  10x , x 2  60, x  60  2 15  7.75
c. In a right triangle, the altitude from the right angle forms three geometric means:
 Starts at the BRA (Big Right Angle) – the three lines coming from there are
y
m
your geometric means: m  m , k  k , y  y
k
 If you go down m, f and h are going to tackle you
 If you go down k, f and g are going to tackle you
 If you go down y, g and h are going to tackle you
Parallel lines cut out proportional parts
Ex: 6y  34 , x x 3  34
f
g
h

6
14
24

x
18
a

d
b

c

Ex: Prove if
J
K
Ex:
x+3
4
Proportional Parts in  s: Sometimes helpful to separate  s
14  x
x
14+x
24

x
18
24
18
Angle Bisector Theorem: An angle bisector cuts the opposite side into proportional parts
a
d
b  c
Similar figures have a proportional perimeter, area is proportional by the scale factor squared
and volume is proportional by the scale factor cubed.
Three ways to prove triangles similar:
 AA ~: If two sets of corresponding angles are  then the  s are ~
 SAS~: If two sets of corresponding sides are proportional and the included
angles are  then the  s are ~
 SSS~: If three sets of corresponding sides are proportional then the  s are ~
FG  FK , KH  FK then JFG  JKH
Proof:
JFG, JKH are rt s
JFG  JKH
FG  FK , KH  FK
Given
F
3
x
y
G
H
 lines form rt s
FJG  KJH
Reflexive
rt s are 
JFG ~ JKH
AA~
Given: BH TL , BH = 2x, TL = x, and T is the midpoint of EH Prove: BHE ~ LTH
Proof: 1) BH TL , BH = 2x, TL = x, and T is the midpoint of EH
TL
 2xx  12
2) BH
3) TH is ½ of EH
TL
 TH
4) BH
EH
5) BHE  LTH
6) BHE ~ LTH
1) Given
H
B
2) Division
3) Def of midpoint
4) Substitution
T
E
5) If lines are II then alt. int. s are 
6) SAS ~
L
Exercises:
x
5

2 x3
1.) Solve for x:
2.) Find the geometric mean between 14 and 21
3.) Solve for x, y, and z.
4
3
x
2
z
3
4.) Solve for x and y.
20
x
y
12
5.) Solve for x.
x
4
x+3
4
6
6.) Two rectangular prisms are similar. The smaller has a side of 4 cm and the corresponding side on the larger prism
is 12 cm. If the volume of the smaller prism is 24 cm3, find the volume of the larger prism.
7.) Given the following picture, prove that BAC ~ JOY
B
6x
J
A
12
4x
O
8
y+y
Y
y+y+y
C
8.) Prove that if B is the midpoint of AC and D is the midpoint of CE then BCD ~ ACE
A
B
C
D
E
9.) Prove that if GF  FH and HC  FH , then GFJ ~ CHJ
G
H
F
J
C
VII. Chapter 9: Right Triangle Trigonometry
 Pythagorean Theorem
a. a 2  b 2  c 2 if a and b are the legs and c is the hypotenuse of a rt. 
x+5
b. Sometimes it is easiest to find the whole missing side first and then subtract:
4
8
Ex: 4 2  8 2  c 2  16  64  c 2  80  c 2  8.9  c  8.9  x  5  3.9  x
 Special Right Triangles
a. Isosceles Right Triangles
 Angles have measures of 45, 45, and 90 degrees
 Half of a square
y
x
45
 Sides have a ratio of 1 : 1 : 2
3
10
Ex1: The legs are equal and the hypotenuse is
Ex2: To go from the hypotenuse to a leg you divide by 2 and the
legs are equal. Not forgetting to reduce we get x  y  5 2
2 times the leg therefore x  3, y  3 2
y
45
x
Ex3: To find the area of the square given that the diagonal is 11 ft: the area of a square is length x width where the length
and width are equal. To go from the hypotenuse to a leg we divide by
2 and so we get a side to be
11 2
2
11
ft long. Squaring
this we get 60.5 ft2.
30
y
x
2
b. 30-60-90 Triangles
 Half of an equilateral triangle
 Sides have a ratio of 1 : 3 : 2
Ex1: To go from short leg to long leg you multiply by 3 ,
Ex2: To go from long leg to short leg you divide
to go from short leg to hypotenuse you multiply by 2.
by
y
60 x
5
3 and then from short leg to hypotenuse
you multiply by 2. Therefore x 
Therefore x  2 3 , y  4
5 3
3
, y  103 3 .
Ex3: To find the perimeter of an equilateral triangle given the altitude is 12cm: to go from the long leg to the short leg you divide
3 . Not forgetting to reduce and then multiplying by 2 to get the hypotenuse we get a side of the equilateral triangle
12
to be 8 3 cm. Multiplying by three to get the perimeter we get 24 3 cm.
 Right Triangle Trigonometry (CHECK TO ENSURE YOUR CALC. IS IN DEGREES!)
a. Sine (sin): the sine of an  in a rt.  is the ratio of the opposite to the hypotenuse
b. Cosine (cos): the cosine of an  in a rt.  is the ratio of the adjacent side to the hyp.
x
c. Tangent (tan): the tangent of an  in a rt.  is the ratio of the opp side to the adj side.
d. Pneumonic devices: SOHCAHTOA, “Some old hippie caught another hippie tripping
36
on acorns”, “some old horse caught another horse taking oats away”

18
sin 36   x
y
tan 41  12z
cos 67   42y
by
18
Ex.
 
x  sin 36  18
x
18
sin 36
 
x  30.6

Ex.
 
42  cos67   y

67
y  16.4
e. Inverse functions: whenever the variable is the angle you
need to use the inverse trig function to solve for the angle
42
 
Ex. 12  tan 41   z
z

41
z  10.4
12
Ex: Dealing with the adj
and the hyp so use cos.
cos x   14
23
23
x
Angle of Elevation and Depression

x  cos 1 14
14
23
a. Angle of Elevation is angle up from the horizontal

x  52.5
b. Angle of Depression is down from the horizontal
c. Since the lines are parallel and Angle of Elevation and Angle of Depression are alt.
int. angles, Angle of Elevation = Angle of Depression
d. Tips:
 Draw a good picture
 Watch out for extra pieces like the height of a person measuring the angle
Exercises:
1) A boat in calm seas travels in a straight line and ends the trip 37 km east and 38 km south of its
original position. Find the distance of the trip to the nearest tenth of a kilometer.
2) Solve for x.
3) In ΔABC, ∠A is a right angle and m∠B = 60. If AB = 20 ft, find BC
4) Find the area of a square with a diagonal of 12 cm.
5) Find the perimeter of an equilateral triangle with an altitude of 15 ft.
6) A large flagpole is 193 feet tall. On a particular day at noon it casts a 159-foot shadow. What is the
sun's angle of elevation at that time? Round to the nearest tenth.
7) Solve for x.
8) Solve for x.
9) Solve for x.