Geometry
Midterm Review
Segment Addition Postulate
 If B is between A and C, then AB + BC = AC
 (Converse): If AB + BC = AC, then B is between A and C.
AC
A
B
C
Application of Segment Addition Postulate:
Use the Diagram to find KL
38
J
15
K
L
JL = JK + KL
Segment Addition Postulate
38 = 15 + KL
Substitute 38 and 15
23 = KL
Simple Algebra will give you a
solution 23
Bisectors or Midpoints
 Midpoint
 A point that splits a
segment into to equal
halves
 Bisectors
 Segment: A line or a Ray
that passes through the
Midpoint of a segment
 Angle: A line or a ray that
cuts an angle in half
Find Segment Lengths
1. M is the midpoint of AB, find AM and MB.
Solution:
M is the midpoint of AB, so AM is half of AB.
AM = ½ AB = ½ 26 = 13
MB = AM = 13
A
M
26
B
Find Segment Lengths
1. P is the midpoint of RS, find PS and RS.
Solution:
P is the midpoint of RS, so PS = RP = 7.
RS = 2 RP = 2 7 = 14
PS = 7 and RS = 14
R
P
S
7
Using Algebra
1.
Line d is a segment bisector of
AB, find x.
Solution:
AM = MB
M is the midpoint, write an equation
5x = 35
Substitute values for AM and MB
X=7
Solve for x
A
M
5x
B
d
35
Laws of Logic
 Law of Detachment
 If the Hypothesis of a statement is true, then the
conclusion is also true.
 Law of Syllogism (aka The Chain Rule)
 If the hypothesis (p), then the conclusion (q)
 If the hypothesis (q), then the conclusion (r)
 If the hypothesis (p), then the conclusion (r)
The Law of Detachment
 Mary goes to the movies every
Friday and Saturday. Today is
Friday
 1st Identify the hypothesis and
conclusion of the statement
 Hypothesis:
 “If it is Friday or Saturday”
 Conclusion:
 “Then Mary will go to the
movies.”
 “Today is Friday” satisfies
the hypothesis, so you can
conclude that Mary will go
to the movies.
The Law of Syllogism
 If Ron gets lunch today,
then he will get a
sandwich.
 If Ron gets a sandwich,
then he will get a glass of
milk.
 If p, then q
 If q, then r
 If Ron gets lunch today,
then he will get a glass of
milk.
 If p, then r
Types of Logical Statements
 If it is raining, then it is
cloudy.
 If it is cloudy, then it is
raining.
 If it is not raining, then it is
not cloudy.
 If it not cloudy, then it is
not raining.
 Conditional Statement:
 Converse:
p®q
q® p
 Inverse:
Øp®Øq
 Contrapositive:
Øq ®Øp
Angles Formed by Transversals
 Corresponding Angles:
 Two angles that are in
corresponding positions
on both the transversal
and accompanying lines
t
1
m
5

1 & 5 are to the left
of the transversal and on
the top of their
accompanying lines
n
Angles Formed by Transversals
 Alternate Interior Angles:
 Two angles that are on the
opposite sides of the
transversal and lie
between the two
accompanying lines
t
m
3
6

3 & 6 are on opposite or
alternating sides of the
transversal and lie on the
inside of the two
accompanying lines
n
Angles Formed by Transversals
 Alternate Exterior Angles:
 Two angles that are on the
opposite sides of the
transversal and lie on the
outside of accompanying
lines
t
2
m
n

2 & 7 are on opposite or
alternating sides of the
transversal and lie on the
outside of the two
accompanying lines
7
Angles Formed by Transversals
 Consecutive Interior Angles:
(AKA Same Side Interior Angles)
 Two angles that are on the
same side of the transversal
and lie between the two
accompanying lines
t
m
4
6

4 & 6 are on the same side
of the transversal and lie on
the inside of the two
accompanying lines
n
Properties of Slope
 Slope:
 Rise/Run
 (y2 – y1)/(x2 – x1)
 Negative Slope
 Moves down from left to right
 Positive Slope
 Moves up from left to right
 Undefined Slope
 Slope of Vertical Lines, y/0
 Zero Slope
 Slope of Horizontal Lines, 0/x
Identify the Parallel Lines
 Which of the lines if any are
parallel?
(2, 2)
(4, 2)
 Slope of p:
 (-6 – (-1))/(-4 – (-3))
 -5/-1 = 5
 Slope of h:
(-3, -1)
(3, -3)
(1, -4)
 (2 – (-4))/(2 – 1)
 6/1 = 6
 Slope of s:
 (2 – (-3))/(4 – 3)
 5/1 = 5
  p  s
p
(-4, -6)
h
s
Slopes of Perpendicular Lines
 Two nonvertical lines are perpendicular if and only if
the product of their slopes is -1
 In other words the slopes of perpendicular lines are
opposite reciprocals
 Example: (5/4)(-4/5) = -1
 Horizontal lines are perpendicular to vertical lines
Drawing a Perpendicular Line
 Line w passes through (1, -2) and (5,
6). Graph the line perpendicular to
line w that passes through (2, 5)
w
(5, 6)
(2, 5)
 Step 1: Find the slope of w
(4, 4)
 (6 – (-2))/(5 – 1) = 8/4 = 2
 Step 2: Determine the slope of the
line perpendicular to w

m = - ½
 Step 3: Use rise and run to find a
second point on the line
(1, -2)
Parts of a Right Triangle
 Hypotenuse
 Longest side of a right
triangle
 Label the Hypotenuse and the
legs of the below Triangle
B
 Side opposite the right
angle
A
 Legs of a Right Triangle
 Two shorter legs of a right
triangle
 Hypotenuse: BC
 The two legs that make up
the right angle
 Legs: AB & AC
C
Using the Pythagorean Theorem
to find…
The Hypotenuse
One of the legs
 Hypotnuse2 = (leg1)2 + (leg2)2
 Hypotnuse2 = (leg1)2 + (leg2)2
 c2 = 3 2 + 4 2
 102 = 62 + b2
 c2 = 9 + 16
 100 = 36 + b2
 c2 = 25
 b2 = 64
 c=5
 b=8
c
3
10
6
4
b
Classifying Triangles using the
Pythagorean Theorem
Acute
Obtuse
 If the sum of the squares of the
two shorter sides is greater than
the square of the largest side,
then the triangle is acute
7
8
 If the sum of the squares of the
two shorter sides is less than the
square of the largest side, then
the triangle is obtuse
6
10
9
12
 72 + 82 ? 102
 62 + 92 ? 122
 49 + 64 ? 100
 36 + 81 ? 144
 113 > 100
 117 < 144
 Therefore the Triangle is Acute
 Therefore the Triangle is Obtuse
Classifying Triangles by their
Sides
 Scalene Triangle
 Equilateral Triangle
No Congruent Sides
 Isosceles Triangle
3 Congruent Sides
At Least 2 Congruent Sides
Classifying Triangles by Angles
 Acute Triangle
 Obtuse Triangle
3 Acute Angles
1 Obtuse Angle
 Right Triangle
1 Right Angle
 Equiangular Triangle
3 Congruent Angles
Interior Angles of a Triangle
Triangle Sum Theorem
Corollary to the Triangle Sum
Theorem
 The sum of the measures of the
angles of a triangle is 180°
 The Acute angles of a right
triangle are complementary
 mA + mB + mC = 180
 mB + mC = 90
B
A
B
C
A
C
Exterior Angle Theorem
 The measure of the exterior angle of a triangle is equal
to the sum of the measures of the two nonadjacent or
opposite angles
 m1 = mA + mB
B
A
1
Triangle Inequalities
 If one side of a triangle is longer than another, then the
angle opposite the longer side is larger than the angle
opposite the shorter side.
 If
AC > AB ,
then
ÐB > ÐC
A
 The converse is also true
B
C
Midsegment
 Properties of a Midsegment
 BD is a Midsegment
 Segment that connects the
midpoints of two sides of a
triangle
C
B
 The Midsegment is half the
length of the third side
 The Midsgment is parallel to the
third side
D
A
 BD = ½ (AE)
 If AE = 12, then BD = 6
E
Medians and Centroids
 A Median connects a vertex of a
triangle to a midpoint of the
opposite side
 P is a Centroid
 The intersection of three
Medians is a Centroid
 The distance from the vertex to
the Centroid is two-thirds the
length of the Median
P
X
A
 AX is a Median
 AP = (2/3)(AX)
 If AX = 27, then AP = 18