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USING UNDEFINED TERMS AND DEFINITIONS
A point has no dimension.
It is usually represented by a small dot.
Point A
A
•
USING UNDEFINED TERMS AND DEFINITIONS
A line extends in one dimension. It is usually represented
by a straight line with two arrowheads to indicate that
the line extends without end in two directions.
Line
or AB
Collinear points are points that lie on the same line.
USING UNDEFINED TERMS AND DEFINITIONS
A plane extends in two dimensions. It is
usually represented by a shape that looks
like a table or a wall, however you must
imagine that the plane extends without end.
USING UNDEFINED TERMS AND DEFINITIONS
Coplanar points are points that lie on the same plane.
A
B
C
M
Plane M or plane ABC
Naming Collinear and Coplanar Points
• Name three points that are collinear.
H
• Name four points that are coplanar.
G
• Name three points that are not collinear.
E
F
D
SOLUTION
• Points D, E, F lie on the same line, so they are collinear.
• Points D, E, F, and G lie on the same plane, so they are coplanar.
Also, D, E, F, and H are coplanar.
• There are many correct answers. For instance, points H, E, and G do
not lie on the same line.
USING UNDEFINED TERMS AND DEFINITIONS
Another undefined concept in geometry is the idea that
a point on a line is between two other points on the line.
You can use this idea to define other important terms
in geometry.
USING UNDEFINED TERMS AND DEFINITIONS
Consider the line AB (symbolized by AB).
The line segment or segment AB
(symbolized by AB) consists of the
endpoints A and B, and all points on AB
that are between A and B.
USING UNDEFINED TERMS AND DEFINITIONS
The ray AB (symbolized by AB) consists
of the initial point A and all points on AB
that lie on the same side of A as point B.
Note that AB is the same as BA, and AB
is the same as BA.
However, AB and BA are not the same.
They have different initial points and
extend in different directions.
USING UNDEFINED TERMS AND DEFINITIONS
If C is between A and B, then CA
and CB are opposite rays.
Like points, segments and rays are collinear if they lie
on the same plane. So, any two opposite rays are
collinear. Segments, rays, and lines are coplanar if they
lie on the same plane.
Drawing Lines, Segments, and Rays
Draw three noncollinear points J, K, L.
Then draw JK, KL and LJ.
SOLUTION
1
Draw J, K, and L
2
Draw JK.
3
Draw KL.
4
Draw LJ.
K
J
L
Drawing Opposite Rays
Draw two lines. Label points on the lines and name two
pairs of opposite rays.
SOLUTION
Points M, N, and X are
collinear and X is between
M and N.
So, XM and XN are
opposite rays.
Points P, Q, and X are collinear and X is between P and Q.
So, XP and XQ are opposite rays.
SKETCHING INTERSECTIONS OF LINES AND PLANES
Two or more geometric figures intersect if they have one
or more points in common. The intersection of the figures
is the set of points the figures have in common.
Sketching Intersections
Sketch a line that intersects a plane at one point.
SOLUTION
Draw a plane and a line.
Emphasize the point
where they meet.
Dashes indicate where
the line is hidden by
the plane.
Sketching Intersections
Sketch two planes that intersect in a line.
SOLUTION
Draw two planes.
Emphasize the line where
they meet.
Dashes indicate where
one plane is hidden
by the other plane
CONGRUENCE OF ANGLES
THEOREM
THEOREM 2.2 Properties of Angle Congruence
Angle congruence is r ef lex ive, sy mme tric, and transitive.
Here are some examples.
REFLEX IVE
For any angle A,
SYMMETRIC If
A 
B, then
TRANSITIVE If
A 
B and
B 
C, then
A 
A
B 
A
A 
C
Transitive Property of Angle Congruence
Prove the Transitive Property of Congruence for angles.
SOLUTION
To prove the Transitive Property of Congruence for angles,
begin by drawing three congruent angles. Label the vertices as
A, B, and C.
A
B
GIVEN
B,
C
PROVE
A
C
B
A
C
Transitive Property of Angle Congruence
A
B
GIVEN
B,
C
Statements
A 
B 
1
PROVE
A
C
Reasons
B,
C
Given
2
m
A=m
B
Definition of congruent angles
3
m
B=m
C
Definition of congruent angles
4
m
A=m
C
Transitive property of equality
5
A 
C
Definition of congruent angles
Using the Transitive Property
This two-column proof uses the Transitive Property.
GIVEN
m
3 = 40°,
PROVE
m
1 = 40°
1
2,
Statements
1
2
m
1
3
3
m
1=m
4
m
1 = 40°
3
Reasons
3 = 40°,
2
3
1
2
2,
Given
Transitive property of Congruence
3
Definition of congruent angles
Substitution property of equality
Proving Theorem 2.3
THEOREM
THEOREM 2.3 Right Angle Congruence Theorem
All right angles are congruent.
You can prove Theorem 2.3 as shown.
GIVEN
1 and
PROVE
1
2 are right angles
2
Proving Theorem 2.3
GIVEN
1 and
PROVE
1
2 are right angles
2
Statements
1 and
1
Reasons
2 are right angles
2
m
1 = 90°, m
3
m
1=m
4
1
2
2
2 = 90°
Given
Definition of right angles
Transitive property of equality
Definition of congruent angles
PROPERTIES OF SPECIAL PAIRS OF ANGLES
THEOREMS
THEOREM 2.4 Congruent Supplements Theorem
If two angles are supplementary to the same angle (or to
congruent angles) then they are congruent.
1
2
3
PROPERTIES OF SPECIAL PAIRS OF ANGLES
THEOREMS
THEOREM 2.4 Congruent Supplements Theorem
If two angles are supplementary to the same angle (or to
congruent angles) then they are congruent.
1
2
3
If m 1 + m 2 = 180° and
m 2 + m 3 = 180° then
1
3
1
PROPERTIES OF SPECIAL PAIRS OF ANGLES
THEOREMS
THEOREM 2.5 Congruent Complements Theorem
If two angles are complementary to the same angle (or to
congruent angles) then the two angles are congruent.
5
4
6
PROPERTIES OF SPECIAL PAIRS OF ANGLES
THEOREMS
THEOREM 2.5 Congruent Complements Theorem
If two angles are complementary to the same angle (or to
congruent angles) then the two angles are congruent.
4
5
4
If m 4 + m 5 = 90° and
m 5 + m 6 = 90° then
4
6
6
Proving Theorem 2.4
GIVEN
1 and 2 are supplements
3 and 4 are supplements
1
4
PROVE
2
3
Statements
1 and 2 are supplements
3 and 4 are supplements
1 4
1
2
Reasons
m
m
1+m
3+m
2 = 180°
4 = 180°
Given
Definition of supplementary angles
Proving Theorem 2.4
GIVEN
1 and 2 are supplements
3 and 4 are supplements
1
4
PROVE
2
Statements
3
Reasons
3
m
m
1+m
3+m
2=
4
Transitive property of equality
4
m
1=m
4
Definition of congruent angles
5
m
m
1+m
3+m
2=
1
Substitution property of equality
Proving Theorem 2.4
GIVEN
1 and 2 are supplements
3 and 4 are supplements
1
4
PROVE
2
Statements
6
7
m
2=m
2
3
3
Reasons
3
Subtraction property of equality
Definition of congruent angles
PROPERTIES OF SPECIAL PAIRS OF ANGLES
POSTULATE
POSTULATE 12 Linear Pair Postulate
If two angles form a linear pair, then they are supplementary.
m
1+m
2 = 180°
Proving Theorem 2.6
THEOREM
THEOREM 2.6 Vertical Angles Theorem
Vertical angles are congruent
1
3,
2
4
Proving Theorem 2.6
GIVEN
PROVE
5 and
6 and
5
6 are a linear pair,
7 are a linear pair
7
Statements
Reasons
1
5 and
6 and
6 are a linear pair,
7 are a linear pair
Given
2
5 and
6 and
6 are supplementary,
7 are supplementary
Linear Pair Postulate
3
5
7
Congruent Supplements Theorem
PROPERTIES OF PARALLEL LINES
POSTULATE
POSTULATE 15 Corresponding Angles Postulate
If two parallel lines are cut by a transversal,
then the pairs of corresponding angles
are congruent.
1
2
1
2
PROPERTIES OF PARALLEL LINES
THEOREMS ABOUT PARALLEL LINES
THEOREM 3.4 Alternate Interior Angles
If two parallel lines are cut by a transversal,
then the pairs of alternate interior angles are
congruent.
3
4
3
4
PROPERTIES OF PARALLEL LINES
THEOREMS ABOUT PARALLEL LINES
THEOREM 3.5 Consecutive Interior Angles
If two parallel lines are cut by a transversal,
then the pairs of consecutive interior angles are
supplementary.
5
6
m
5+m
6 = 180°
PROPERTIES OF PARALLEL LINES
THEOREMS ABOUT PARALLEL LINES
THEOREM 3.6 Alternate Exterior Angles
If two parallel lines are cut by a transversal,
then the pairs of alternate exterior angles are
congruent.
7
8
7
8
PROPERTIES OF PARALLEL LINES
THEOREMS ABOUT PARALLEL LINES
THEOREM 3.7 Perpendicular Transversal
If a transversal is perpendicular to one of two parallel
lines, then it is perpendicular to the other.
j
k
Proving the Alternate Interior Angles Theorem
Prove the Alternate Interior Angles Theorem.
SOLUTION
GIVEN
p || q
PROVE
1
Statements
2
Reasons
1
p || q
Given
2
1
3
3 2
Vertical Angles Theorem
4
1
Transitive property of Congruence
3
2
Corresponding Angles Postulate
Using Properties of Parallel Lines
Given that m 5 = 65°,
find each measure. Tell
which postulate or theorem
you use.
SOLUTION
m
6 = m
5 = 65°
m
7 = 180° – m
m
8 = m
5 = 65°
Corresponding Angles Postulate
m
9 = m
7 = 115°
Alternate Exterior Angles Theorem
Vertical Angles Theorem
5 = 115° Linear Pair Postulate
PROPERTIES OF SPECIAL PAIRS OF ANGLES
Using Properties of Parallel Lines
Use properties of
parallel lines to find
the value of x.
SOLUTION
m
m
4 = 125°
4 + (x + 15)° = 180°
125° + (x + 15)° = 180°
x = 40°
Corresponding Angles Postulate
Linear Pair Postulate
Substitute.
Subtract.
Estimating Earth’s Circumference: History Connection
Over 2000 years ago
Eratosthenes estimated Earth’s
circumference by using the
fact that the Sun’s rays are
parallel.
When the Sun shone exactly
down a vertical well in Syene,
he measured the angle the
Sun’s rays made with a
vertical stick in Alexandria.
He discovered that
m
2
1
50 of a circle
Estimating Earth’s Circumference: History Connection
m
2
1
50 of a circle
Using properties of parallel
lines, he knew that
m
1= m
2
He reasoned that
m
1
1
50 of a circle
Estimating Earth’s Circumference: History Connection
m
1
1
50 of a circle
The distance from Syene to
Alexandria was believed to be
575 miles
1
50 of a circle
Earth’s
circumference
575 miles
Earth’s circumference
50(575 miles)
Use cross product property
29,000 miles
How did Eratosthenes know that m
1=m
2?
Estimating Earth’s Circumference: History Connection
How did Eratosthenes know that m
1=m
SOLUTION
Because the Sun’s rays are parallel,
Angles 1 and 2 are alternate interior
angles, so
1 
2
By the definition of congruent angles,
m
1=m
2
2?
SSS AND SAS CONGRUENCE POSTULATES
If all six pairs of corresponding parts (sides and angles) are
congruent, then the triangles are congruent.
If
Sides are
congruent
and
Angles are
congruent
1. AB
DE
4.
A
D
2. BC
EF
5.
B
E
3. AC
DF
6.
C
F
then
Triangles are
congruent
 ABC
 DEF
SSS AND SAS CONGRUENCE POSTULATES
POSTULATE
POSTULATE 19 Side - Side - Side (SSS) Congruence Postulate
If three sides of one triangle are congruent to three sides
of a second triangle, then the two triangles are congruent.
If Side
S MN
QR
Side
S NP
RS
Side
S PM
SQ
then  MNP
 QRS
SSS AND SAS CONGRUENCE POSTULATES
The SSS Congruence Postulate is a shortcut for proving
two triangles are congruent without using all six pairs
of corresponding parts.
Using the SSS Congruence Postulate
Prove that
 PQW
 TSW.
SOLUTION
Paragraph Proof
The marks on the diagram show that PQ  TS,
PW  TW, and QW  SW.
So by the SSS Congruence Postulate, you know that
 PQW   TSW.
SSS AND SAS CONGRUENCE POSTULATES
POSTULATE
POSTULATE 20 Side-Angle-Side (SAS) Congruence Postulate
If two sides and the included angle of one triangle are
congruent to two sides and the included angle of a
second triangle, then the two triangles are congruent.
If
Side
S
Angle
A
Side
S
PQ
WX
Q
X
QS
XY
then  PQS
WXY
Using the SAS Congruence Postulate
Prove that
 AEB DEC.
1
2
3
1
Statements
Reasons
AE  DE, BE  CE
Given
1 2
 AEB   DEC
2
Vertical Angles Theorem
SAS Congruence Postulate
MODELING A REAL-LIFE SITUATION
Proving Triangles Congruent
You are designing the window shown in the drawing. You
want to make  DRA congruent to  DRG. You design the window so that
DR AG and RA  RG.
ARCHITECTURE
Can you conclude that  DRA   DRG ?
D
SOLUTION
GIVEN
PROVE
DR
AG
RA
RG
 DRA
A
 DRG
R
G
Proving Triangles Congruent
GIVEN
DR
AG
RA
RG
 DRA
PROVE
D
 DRG
A
Statements
R
G
Reasons
Given
1
DR
AG
2
DRA and DRG
are right angles.
If 2 lines are , then they form
4 right angles.
3
DRA 
4
RA  RG
Given
5
DR  DR
Reflexive Property of Congruence
6
 DRA   DRG
SAS Congruence Postulate
DRG
Right Angle Congruence Theorem
Congruent Triangles in a Coordinate Plane
Use the SSS Congruence Postulate to show that  ABC   FGH.
SOLUTION
AC = 3 and FH = 3
AC  FH
AB = 5 and FG = 5
AB  FG
Congruent Triangles in a Coordinate Plane
Use the distance formula to find lengths BC and GH.
d=
BC =
(x 2 – x1 ) 2 + ( y2 – y1 ) 2
(– 4 – (– 7)) 2 + (5 – 0 ) 2
d=
GH =
(x 2 – x1 ) 2 + ( y2 – y1 ) 2
(6 – 1) 2 + (5 – 2 ) 2
=
32 + 52
=
52 + 32
=
34
=
34
Congruent Triangles in a Coordinate Plane
BC = 34 and GH = 34
BC  GH
All three pairs of corresponding sides are congruent,
 ABC   FGH by the SSS Congruence Postulate.
USING MEDIANS OF A TRIANGLE
A median of a triangle is a
segment whose endpoints are
a vertex of the triangle and the
midpoint of the opposite side.
D
USING MEDIANS OF A TRIANGLE
The three medians of a triangle are concurrent. The point
of concurrency is called the centroid of the triangle. The
centroid is always inside the triangle.
centroid
acute triangle
centroid
centroid
right triangle
obtuse triangle
The medians of a triangle have a special concurrency property.
USING MEDIANS OF A TRIANGLE
THEOREM
THEOREM 5.7 Concurrency of Medians of a Triangle
The medians of a triangle intersect at a point that is two-thirds
of the distance from each vertex to the midpoint of the
opposite side.
If P is the centroid of ABC, then
2
AD
3
BP = 2 BF
3
CP = 2 CE
3
AP =
P
USING MEDIANS OF A TRIANGLE
The centroid of a triangle
can be used as its
balancing point.
A triangular model of uniform
thickness and density will
balance at the centroid of the triangle.
Using the Centroid of a Triangle
P is the centroid of QRS shown below and PT = 5.
Find RT and RP.
SOLUTION
Because P is the centroid,
RP = 2 RT.
3
1
Then PT = RT – RP = RT
3
1
Substituting 5 for PT, 5 =
RT, so RT = 15.
3
Then RP =
2
RT = 2 (15) = 10.
3
3
So, RP = 10 and RT = 15.
Finding the Centroid of a Triangle
Find the coordinates of the centroid of  JKL.
(5, 8)
SOLUTION
The centroid is two thirds of the
distance from each vertex to the
midpoint of the opposite side.
Choose the median KN. Find
the coordinates of N, the midpoint
of JL.
The coordinates of N are
3 + 7 , 6 + 10
2
2
L
= (5, 8)
(7, 10)
N
P
(5, 2)
K
=
10
, 16
2
2
(3, 6)
J
Finding the Centroid of a Triangle
Find the coordinates of the centroid of  JKL.
(5, 8)
SOLUTION
Find the distance from vertex K
to midpoint N. The distance from
K (5,2) to N(5,8) is 8 – 2, or 6 units.
Determine the coordinates of
2
the centroid, which is • 6, or
3
4 units up from vertex K along
(3, 6)
(7, 10)
N
(5, 6)
P
L
J
M
(5, 2)
K
the median KN.
The coordinates of the centroid P are (5, 2 + 4), or (5, 6)
[Yellow coordinates appear.]
USING ALTITUDES OF A TRIANGLE
An altitude of a triangle is the perpendicular segment
from a vertex to the opposite side or to the line that contains
the opposite side. An altitude can lie inside, on, or outside
the triangle.
Every triangle has three altitudes. The lines containing the
altitudes are concurrent and intersect at a point called
the orthocenter of the triangle.
Drawing Altitudes and Orthocenters
Where is the orthocenter of an acute triangle?
SOLUTION
Draw an example.
The three altitudes intersect at G,
a point inside the triangle.
Drawing Altitudes and Orthocenters
Where is the orthocenter of a right triangle?
SOLUTION
The two legs, LM and KM,
are also altitudes. They intersect
at the triangle’s right angle.
This implies that the orthocenter is on the triangle
at M, the vertex of the right angle of the triangle.
Drawing Altitudes and Orthocenters
Where is the orthocenter of an obtuse triangle?
SOLUTION
The three lines that contain the
altitudes intersect at W,
a point outside the triangle.
USING ALTITUDES OF A TRIANGLE
THEOREM
THEOREM 5.8 Concurrency of Altitudes of a Triangle
The lines containing the altitudes of a triangle
are concurrent.
If AE, BF, and CD are the altitudes
of ABC, then the lines AE, BF, and
CD intersect at some point H.
PROPERTIES OF SPECIAL PARALLELOGRAMS
A rhombus is a parallelogram
with four congruent sides.
A rectangle is a parallelogram
with four right angles.
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|
|
|
|
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A square is a parallelogram with four congruent sides
and four right angles.
PROPERTIES OF SPECIAL PARALLELOGRAMS
The Venn diagram shows the relationships among
parallelograms, rhombuses, rectangles, and squares.
Each shape has the properties of every group that it
belongs to. For instance, a square is a rectangle, a
rhombus, and a parallelogram, so it has all of the
properties of each of those shapes.
PROPERTIES OF SPECIAL PARALLELOGRAMS
parallelograms
rhombuses
rectangles
squares
Describing a Special Parallelogram
Decide whether the statement is always, sometimes, or never true.
A rhombus is a rectangle.
SOLUTION
The statement is sometimes true.
In the Venn Diagram, the regions for rhombuses and
rectangles overlap. If the rhombus is a square, it is a
rectangle.
Help
Describing a Special Parallelogram
Decide whether the statement is always, sometimes, or never true.
A parallelogram is a rectangle.
SOLUTION
The statement is sometimes true.
Some parallelograms are rectangles. In the Venn
diagram, you can see that some of the shapes in the
parallelogram box are in the region for rectangles, but
many aren’t.
Help
Using Properties of Special Parallelograms
ABCD is a rectangle. What else do you know about ABCD?
A
B
D
C
SOLUTION
Because ABCD is a rectangle, it has four right angles by the definition.
The definition also states that rectangles are parallelograms, so ABCD
has all the properties of a parallelogram:
1
Opposite sides are parallel and congruent.
2
Opposite angles are congruent and consecutive angles
are supplementary.
3
Diagonals bisect each other.
Using Properties of Special Parallelograms
ABCD is a rectangle. What else do you know about ABCD?
A
B
D
C
A rectangle is defined as a parallelogram with four
right angles. But any quadrilateral with four right
angles is a rectangle because any quadrilateral with
four right angles is a parallelogram.
Using Properties of Special Parallelograms
COROLLARIES ABOUT SPECIAL QUADRILATERALS
RHOMBUS COROLLARY
A quadrilateral is a rhombus if and only if it has four congruent sides.
RECTANGLE COROLLARY
A quadrilateral is a rectangle if and only if it has four right angles.
SQUARE COROLLARY
A quadrilateral is a square if and only if it is a rhombus and a rectangle.
You can use these corollaries to prove that a quadrilateral is a rhombus, rectangle,
or square without proving first that the quadrilateral is a parallelogram.
Using Properties of a Rhombus
In the diagram, PQRS is a rhombus.
What is the value of y?
P
Q
2y + 3
S
SOLUTION
5y – 6
All four sides of a rhombus are congruent, so RS = PS.
5y – 6 = 2y + 3
Equate lengths of congruent sides.
5y = 2y + 9
Add 6 to each side.
3y = 9
Subtract 2y from each side.
y=3
Divide each side by 3.
R
USING DIAGONALS OF SPECIAL PARALLELOGRAMS
THEOREMS
THEOREM 6.11
A parallelogram is a rhombus if and only if its
diagonals are perpendicular.
ABCD is a rhombus if and only if AC
BD
USING DIAGONALS OF SPECIAL PARALLELOGRAMS
THEOREMS
THEOREM 6.12
B
C
A parallelogram is a rhombus if and only if each
diagonal bisects a pair of opposite angles.
A
ABCD is a rhombus if and only if AC bisects
BD bisects
D
DAB and
ADC and
BCD and
CBA
USING DIAGONALS OF SPECIAL PARALLELOGRAMS
THEOREM S
THEOREM 6.13
A
B
D
C
A parallelogram is a rectangle if and
only if its diagonals are congruent.
ABCD is a rectangle if and only if AC  BD
USING DIAGONALS OF SPECIAL PARALLELOGRAMS
You can rewrite Theorem 6.11 as a conditional statement and its converse.
Conditional statement:
If the diagonals of a parallelogram are
perpendicular, then the parallelogram is
a rhombus.
Converse:
If a parallelogram is a rhombus, then its
diagonals are perpendicular.
To prove the theorem, you must prove both statements.
USING SIMILARITY THEOREMS
THEOREM S
THEOREM 8.2 Side-Side-Side (SSS) Similarity Theorem
If the corresponding sides of two
triangles are proportional, then the
triangles are similar.
P
A
Q
R
If AB = BC = CA
RP
PQ
QR
then ABC ~ PQR.
B
C
USING SIMILARITY THEOREMS
THEOREM S
THEOREM 8.3 Side-Angle-Side (SAS) Similarity Theorem
If an angle of one triangle is
congruent to an angle of a
second triangle and the lengths
of the sides including these
angles are proportional, then the
triangles are similar.
If
X
XY
M and ZX = MN
PM
then XYZ ~ MNP.
M
X
P
Z
Y
N
Proof of Theorem 8.2
GIVEN
RS
= ST = TR
LM
MN
NL
PROVE
 RST ~  LMN
SOLUTION Paragraph Proof
M
L
S
P
N
Q
R
T
Locate P on RS so that PS = LM.
Draw PQ so that PQ
RT.
Then  RST ~  PSQ, by the AA Similarity Postulate, and RS = ST = TR .
PS
SQ
QP
Because PS = LM, you can substitute in the given proportion and
find that SQ = MN and QP = NL. By the SSS Congruence Theorem,
it follows that  PSQ   LMN.
Use the definition of congruent triangles and the AA Similarity Postulate to
conclude that  RST ~  LMN.
Using the SSS Similarity Theorem
Which of the following three triangles are similar?
12
A
E
C
6
6
9
F
14
G
J
4
8
D
6
B
10
H
SOLUTION
To decide which of the triangles are similar, consider the
ratios of the lengths of corresponding sides.
Ratios of Side Lengths of  ABC and  DEF
AB
6
3
=
=
,
DE
4
2
Shortest sides
CA
12
3
=
=
,
FD
8
2
Longest sides
BC
9
3
=
=
EF
6
2
Remaining sides
Because all of the ratios are equal,  ABC ~  DEF
Using the SSS Similarity Theorem
Which of the following three triangles are similar?
12
A
E
C
6
6
9
F
14
G
J
4
8
D
6
B
10
H
SOLUTION
To decide which of the triangles are similar, consider the
ratios of the lengths of corresponding sides.
Ratios
Lengths
of  ABC
and and
GHJ ABC is not
SinceofSide
ABC
is similar
to  DEF
AB
6 to  GHJ,  DEF
CA is12
6
similar
not= similar
=
,
=
=
1
,
GH
6
JG
14
7
Shortest sides
Longest sides
BC .
to  GHJ
=
9
HJ
10
Remaining sides
Because all of the ratios are not equal,  ABC and  DEF are not similar.
Using the SAS Similarity Theorem
Use the given lengths to prove that  RST ~  PSQ.
SOLUTION
GIVEN
SP = 4, PR = 12, SQ = 5, QT = 15
PROVE
 RST ~  PSQ
S
Use the SAS Similarity Theorem.
Find the ratios of the lengths of the corresponding sides.
Paragraph Proof
4
P
SR
SP + PR
4 + 12
16
=
=
=
= 4
SP
SP
4
4
ST
SQ + QT
5 + 15
20
=
=
=
= 4
SQ
SQ
5
5
12
R
The side lengths SR and ST are proportional to the corresponding
side lengths of  PSQ.
Because S is the included angle in both triangles, use the
SAS Similarity Theorem to conclude that  RST ~  PSQ.
5
Q
15
T
USING SIMILAR TRIANGLES IN REAL LIFE
Using a Pantograph
SCALE DRAWING As you move the tracing pin of a pantograph along
a figure, the pencil attached to the far end draws an enlargement.
P
R
Q
T
S
USING SIMILAR TRIANGLES IN REAL LIFE
Using a Pantograph
As the pantograph expands and contracts, the three brads and the
tracing pin always form the vertices of a parallelogram.
P
R
Q
T
S
USING SIMILAR TRIANGLES IN REAL LIFE
Using a Pantograph
The ratio of PR to PT is always equal to the ratio of PQ to PS. Also,
the suction cup, the tracing pin, and the pencil remain collinear.
P
R
Q
T
S
Using a Pantograph
P
R
How can you show that  PRQ ~  PTS?
T
Q
SOLUTION
S
PR
PQ
You know that PT = PS . Because P  P, you can apply the
SAS Similarity Theorem to conclude that  PRQ ~  PTS.
Using a Pantograph
P
In the diagram, PR is 10 inches and
RT is 10 inches. The length of the cat,
RQ,
original
print
2.4 inches.
Find in
thethe
length
TS in
theisenlargement.
10"
R
2.4"
10"
T
Q
SOLUTION
Because the triangles are similar, you can set up a proportion
to find the length of the cat in the enlarged drawing.
PR RQ
=
Write proportion.
PT TS
10 = 2.4
Substitute.
20
TS
Solve for TS.
TS = 4.8
So, the length of the cat in the enlarged drawing is 4.8 inches.
S
Finding Distance Indirectly
ROCK CLIMBING
are at an
indoor
climbing
wall.
To estimate
of
SimilarYou
triangles
can
be used
to find
distances
thatthe
areheight
difficult
the wall, youto
place
a mirror
on the floor 85 feet from the base of the wall. Then
measure
directly.
you walk backward until you can see the top of the wall centered in the mirror.
You are 6.5 feet from the mirror and your eyes are 5 feet above the ground.
Use similar triangles to estimate
the height of the wall.
D
B
5 ft
Not drawn to scale
A
6.5 ft
C
85 ft
E
Finding Distance Indirectly
Use similar triangles to estimate
the height of the wall.
SOLUTION
Due to the reflective property of mirrors,
you can reason that ACB 
ECD.
D
Using the fact that  ABC and  EDC
are right triangles, you can apply the
AA Similarity Postulate to conclude
that these two triangles are similar.
B
5 ft
A
6.5 ft
C
85 ft
E
Finding Distance Indirectly
Use similar triangles to estimate
the height of the wall.
SOLUTION
DE EC
Ratios of lengths of
=
corresponding sides are equal.
BA AC
So,
the height of the wall is about 65 feet.
DE = 85
Substitute.
5
6.5
D
Multiply each side by
5 and simplify.
65.38  DE
B
5 ft
A
6.5 ft
C
85 ft
E