Geometry:
A Complete Course
(with Trigonometry)
Module D Student WorkText
Written by: Thomas E. Clark
Larry E. Collins
Geometry: A Complete Course (with Trigonometry)
Module D–Student Worktext
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Table of Contents
Unit IV - Triangles
Part A - Basic Definitions
LESSON 1
LESSON 2
Triangle Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315
Triangle Types. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 320
Part B - Basic Theorems
LESSON 1
LESSON 2
Theorem 25: “If you have any given triangle, then the sum of the
measures of its angles is 180°.” . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324
Theorem 26 – “If you have a given exterior angle of a triangle, then
its measure is equal to the sum of the measures of the two remote
interior angles.”. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327
Part C - Similarity – Part 1 (General Geometric Relationship)
LESSON 1
LESSON 2
LESSON 3
Ratio and Proportion. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331
Special Properties of Proportions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336
Theorem 27: “If you have two similar polygons, then the ratio of
the perimeters of the two polygons is equal to the ratio of the
corresponding sides.” . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341
Part D - Similarity – Part 2 (Triangles and Their Parts)
LESSON 1
LESSON 2
LESSON 3
LESSON 4
LESSON 5
LESSON 6
Postulate 12 – Triangle Similarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346
Theorem 28: “If a line is parallel to one side of a triangle and
intersects the other two sides in different points, then it divides the
two sides proportionally.” . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351
Theorem 29: “If two triangles are similar, then the measures of
corresponding altitudes are in the same ratio as the measures of
corresponding sides.” . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 358
Theorem 30: “If the altitude is drawn to the hypotenuse of a right
triangle, then the two triangles formed are similar to the original
triangle, and to each other.” . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363
Theorem 31: “If you have a given right
triangle, then the square of the measure of the hypotenuse is equal to
the sum of the squares of the measures of the two legs.”
(The Pythagorean Theorem) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 369
Applying Pythagoras to 3-Dimensional Figures . . . . . . . . . . . . . . . . . . 377
Part E - Congruence – Part 1 (General Geometric Relationship)
LESSON 1
LESSON 2
LESSON 3
Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382
Postulate 13: Triangle Congruence. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 389
Congruence Postulate Corollaries. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 395
Module D - Table of Contents
i
Part F - Congruence – Part 2 (Applications)
LESSON 1
LESSON 2
LESSON 3
LESSON 4
LESSON 5
LESSON 6
LESSON 7
Overlapping Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .400
Using the Definition of Congruence (CPCTC) . . . . . . . . . . . . . . . . . . .404
Theorem 32: “If two given triangles are both congruent to a third
triangle, then the two given triangles are congruent to each other.” . . .414
Theorem 33: “If two sides of a triangle are congruent, then the
angles opposite them are congruent.” . . . . . . . . . . . . . . . . . . . . . . . . . . .415
Theorem 34: “If two angles of a triangle are congruent, then the
sides opposite them are congruent.” . . . . . . . . . . . . . . . . . . . . . . . . . . . .422
Theorem 35: “If a ray bisects an angle of a triangle, then it intersects
the opposite side in such a way that, the two segments formed are in
the same ratio as the two other sides.” (Angle Bisector Theorem) . . . .426
Theorem 36: “If the sum of the squares of the measures of two sides
of a triangle is equal to the square of the measure of the third side,
then the triangle is a right triangle.” (Converse of the Pythagorean
Theorem) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .430
Part G - Congruence – Part 3 (Triangle Inequalities)
LESSON 1
LESSON 2
LESSON 3
LESSON 4
Theorem 37: “If you have a given exterior angle of a triangle, then
the measure of that angle is greater than the measure of either remote
interior angle.” (Exterior Angle Inequality Theorem) . . . . . . . . . . . . . .436
Theorem 38: “In a given triangle, if two sides are not congruent,
then the angles opposite those sides are not congruent.” . . . . . . . . . . . .442
Theorem 39: “In a given triangle, if two angles are not congruent,
then the sides opposite those angles are not congruent.” . . . . . . . . . . . .446
Theorem 40: “In a given triangle, the sum of the lengths of any two
sides, is greater than the length of the third side.” . . . . . . . . . . . . . . . . .451
Appendices
Appendix A - Properties of Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .A-1
Appendix B - Definitions and Important Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .B-1
Appendix C - Postulates and Postulate Corollaries . . . . . . . . . . . . . . . . . . . . . . . . . . . .C-1
Appendix D - Theorems and Theorem Corollaries . . . . . . . . . . . . . . . . . . . . . . . . . . . .D-1
Appendix E - T-Frame Proof Template . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .E-1
Example 1: In each of the following, write the ratios of the two given numbers in lowest
terms.
Solution:
a) 2 to 3
b) 8 to 24
c) 4x : 5x
d) 10x2 to 4x
e) 90 inches to 360 inches
f) 1 meter to 10 meters
g) (x2 + 4x) to (x + 4)
h) 3xy : 6y
a)
c)
e)
2
8
b)
3
4x
5x
=
4•x
5•x
=
90 inches
360 inches
g) x 2 + 4 x
x+4
=
4
5
=
24
2•22•22•11
2
=
2•22•33
22•2
=
1
3
2
• • •
d) 10 x = 22 55 xx xx = 5 x
4x
22•2 •xx
2
2 • 3 • 3 •5
22•22•2•3•33•55
(
x x+4
)= x
( x + 4) 1
•
1
=
1 inch
1 meter
f)
10 meters
4 inches
h) 3 xy
or x
6y
=
33•xx•yy
22•3 •yy
=
x
2
Example 2: In the given figure, AB = 28 and BC = 7. Write the following ratios in lowest
terms.
A
C
B
BC
D
a) AC : AB
b) AB
2x
21
c) BC : AC
27° 24°
9
y
d)
B
x
30°
A
Solution:
a)
c)
332
AC
AB
BC
AC
=
=
21
28
=
3 •7
4 •7 •1
=
7 1•7 1
=
=
21 3 •7 3
Unit IV – Triangles
3
4
12
AB
AC
C
7
BC
7
1
=
b) AB = 28 = 77•2
2•2
2 4
d)
AB
AC
=
28 2
2•7
7 4
2•2
=
=
21
3•7
7
3
3
31. Suppose you ride your bicycle 90km in 3 hours. Compare your distance traveled, to
your riding time, as a fraction. Then, express the fraction as a rate in kilometers per
hour.
32. On a recent weekend trip, Rebecca traveled 437 miles using 23 gallons of gasoline.
Compare the miles Rebecca traveled, to the fuel she used, as a fraction. Then,
express that fraction as a rate, in miles per gallon.
33. Two supplementary angles have a ratio of 2 to 1. What is the measure of each angle?
34. The acute angles of a right triangle are in a ratio of 5 to 4. What is the measure of
each angle?
35. A 56-inch segment is divided in a ratio of 3 : 5. What is the length of each segment?
36. Two numbers are in a ratio of 2 : 3. What is the ratio of their squares?
37. Find the measures of the angles of a triangle, if they are in the ratio of 5 : 7 : 8.
38. The pitch of a roof is the ratio of the rise to the run. If a roof has a rise measuring 1.8
feet, and a run measuring 3.6 feet, what is the pitch? Rewrite this ratio as a decimal to
the nearest hundredth.
Rise
Run
Rise
Run
39. The grade of a residential driveway is the ratio of the
Rise rise to the run. If a driveway has
Run is the grade of the driveway?
a rise of 15 inches and a run of 25 feet, what
Rise
Run
40. If 2x = 3y, find the ratio of x to y.
Part C – Similarity - Part 1 (General Geometric Relationship)
335
Unit IV — Triangles
Part C — Similarity – Part 1 (General Geometric Relationship)
Lesson 2 — Special Properties of Proportions
Objective:
To understand and recognize various special properties of proportions, and
to be able to manipulate proportions using those special properties.
Important Terms:
Geometric Mean in a Proportion – In a proportion, when the second and third
terms are equal in value, that value is called the geometric mean between the
first and fourth terms. Generally:
a
x
If x = d , (with x ? 0, and d ? 0), then x is the geometric mean between
a and d.
Means-Extremes Product Property of a Proportion – In a proportion, the
product of the means, is equal to the product of the extremes. Generally:
If
a
b
c
= d , (with b ? 0, and d ? 0), then a • d = b • c
Equivalent Forms of a Proportion – Recognizing that, in a given valid
proportion, the means-extremes product property is preserved, there are three
additional forms of that proportion which are considered equivalent. Specifically:
If
a
b
c
= d , (with b ? 0, and d ? 0), the following proportions are equivalent forms:
aa bc d c b d
= ,
=
== ,
cb dd b a a c
Denominator-Addition Property of a Proportion – In a proportion, the value of
the denominator of each ratio may be added to the numerator of that ratio
without affecting the integrity of the proportion. Specifically:
If
336
a
b
c
= d, (with b ? 0, and d ? 0), then
Unit IV – Triangles
a+ b c+d
=
b
d
.
Lesson 2 — Exercises:
In Exercises 1-9, find the value of x in each proportion.
1.
13
x
=
24 24
6.
x + 3 3x − 2
=
10
8
2.
14 7
=
x
8
7.
3.
7 9.8
=
12
x
2x − 3 3x − 7
=
3
2
4. 5 : x = 8 : 15
8.
x−3 2
=
2
x
5.
3− x 2
=
x +1 1
9.
x+2
6
=
6
x+2
In Exercises 10-12, make a list for each, using a-g, and then complete each statement in
seven different ways. Give a reason for each answer. (NOTE: Answers will vary, as there
are more than seven ways for each.)
10. If
7 14
=
, then _____________.
8 16
11. If
x 6
= , then _____________.
y 7
12. If
13 19
, then _____________.
=
x
y
a x
=
to complete each proportion.
b y
y
y
= _______.
= _______.
14.
15.
x
b
In Exercises 13-15, use the proportion
13.
a
= _______.
x
c+d u+v
=
to complete each proportion.
d
v
v
c + 2d
17. = _______.
18.
= _______.
u
d
In Exercises 16-18, use the proportion
16.
c
= _______.
d
19. A basketball player attempts 156 shots and makes 117 during one basketball season.
Set up a proportion to determine the player’s shooting percentage, and express that
percentage to the nearest whole percent.
20. The mixture for a finish coat of concrete is one part cement to two parts sand. Set up
a proportion to determine how much cement should be mixed with 15 pounds of sand,
and solve it for the amount of cement needed.
21. A designated hitter for the local minor league baseball team made eight hits in nine
games. If he continues hitting at this rate, how many hits will he make in 108 games?
Part C – Similarity - Part 1 (General Geometric Relationship)
339
22. An old cake recipe requires 2 parts butter to 3 parts sugar. Set up a proportion to
1
determine the amount of butter to be used with 4 2 cups of sugar, and find the amount
of butter needed.
In Exercises 23 and 24, find the geometric mean between the given numbers.
23. 9 and 16
24. 2 and 18
In Exercises 25 and 26, find the fourth proportional for the numbers given.
25. 4, 6, 10
26.
1 1 1
,
,
2 3 4
In Exercises 27-29, find the third proportional for the given pairs of numbers.
a x
(Note: If x = b , then b is often called the third proportional.)
27. 2, 7
28. 1, 9
30. Complete the following generalization: If
29.
1 1
,
5 11
a c e g
a ( a + _____)
= = = = …, then =
.
b d f h
b ( b + _____)
In Exercises 31-33, decide whether each statement is always true, sometimes true, or
never true. Give an explanation for your answer.
31. When two terms of the proportion a : b = c : d are interchanged, the two resulting
ratios form a proportion.
32. If only one term of the proportion a : b = c : d is added to another term, then the two
resulting ratios form a proportion.
33. If
340
a c
a+ b c+ b
= , then
=
.
b d
b
d
Unit IV – Triangles
Example 1: Which of the following triangles are similar to nABC?
E
I
A
5
D
C
60
A
30
E
C
y
G
6x°
A
E
H
30
B
I
J
4x°
2x
OD
R
N
60
x
2x°
B
E
C
6x°
4x°
Q
J
H
60
4
A
6
30
3
x
N
MF
G
L
K
2x
30
5
C
60
A
D
6
F
2x
K
4
A
60
B
3
x
6
D
y
R
P
x
B
D
y
3
9
2x
O
N
nJKL and nPQR have
angles which
areRcongruent to corresponding angles
6
P
M 4 Tmay
x
of nABC, and are, therefore, similar
to nABC. The otherNtriangles
be
M 2x°
6
B information
is needed.
similar to nABC, but more
M
Solution:
8
N
12
Example 2: If /A > /K, /B > /M, and
M
Q
C
A
K
60
P
x
5
87°
E
A
9
AB 5
y
= , find the value of AC when
KN = 12. 5
M 4 T 6 x P J
KM 8
K
1232°
C
2x
N
C
X61°
W
87°
5
8
AC
12
Substituting,
we have61° =
32°
2x
C
B
5
x
H
KG
J
3
x7
F
y
M
D
Z
W
T
y
E
R
10
F
7
F
48°
y
12
18
D
A
A
D
E
R
C
x
8
S
B
B
A
y
7
F
18
C
12
30°
B 20°
N
Q
A
15
16
S
C
D
A
12
Part D –
B
Similarity
A
18
F
B
9
R
O
T
I
8
J
20°
M
A
16
x
8
18
42°30°
Z N
Y
And, 60 = AC or 7P1
O 8
2
C
A
P
So,MX60 = 8•AC
I
8
J
H
D G
x
Then 55! •12
12 =
=8
AC (using the means-extremes property of proportions)
C
8•AC
42°
Y
y
16
10 I
18
KN F
48°
H
y
D
KM
E
A
A
M
12
G
Since nABC z nKMN (Postulate Corollary 12a or 12b),
AB AC
we know that
=
A
3
I
F
61°
B
D
H
3
7
A
61°
K
Solution:
y
12
G
B
8
A
R
5
L
A
B
x
10
B
E
D
C
U
6
-
T
Q 8
Part
16
S
2 (Triangles and A
Their Parts)
15
R
347
18
F
9
x
10
A
J K
F
y
D
D
8
X
12.
Y
A
O
W
18
F
48°
y
Z 7x F
FP
E 7 12
42° E8
A
y
18
188
E 18
12
A
A
M
R
B
F
BD
30°
AE
O
A
15 G
A
16
A
A
9
AC
A
DA
BC
A
R
Prove: H
G
12
U
AB
B
16
6
A
B
16. Given:
S
J
C
IC
U
C
C
E
E
j
C
x
12
x
||
||
C
ED
RC T15
AB BCj
=
ED BADC
6
U C
D8
D F
TS
E x
9
E10 C
L
EA
A
D
D
5D
y
5
D
A
BBE
C
14
B
x
B 9
Q18
AD
AB
D
D
K
E
E
A
C
y
5
10
A
P
S
ADF
350
T
(DE)(AC)
= (AB)(DF)
Q
Unit IV – Triangles
5
6
x
E
B
D
E
E
18
F
A
Ex
E
B
B
A
C
D
E
B
C
A B C
A
A
yA
DC
C
E
FC
F
D
D
B
C
E
F
C
F
E
D
E
B
D
E
F
AB = y
B
RProve:
D
B
11
|| AC; EFB|| BC;C
C B
D
B
C
14
A
E
D
A
AB || DE
BC || EF
17. Given:
x
10
8
T
A
E
18
F
9
D
D
F
21
D
5
C
B
15
18
B A E
E
B
A
17 y
x
11
A
C
D
y
F
7E
1411
E
F
12
E 6 21
C
A
F
17
I
8
J
14.
E
E
14
N
14
x y
y 8
x
5
B
x
B
10
17
y
21
D
S D
A
B
B
E
D
3
6 15
E
I
AD
C8
T
B
15 Q
18 A
F
K
B
9
x R A
16 B 10
P
B
12 E
D
C
8
Q
C
F
D
C
F
F RA
6
6
6
6
T
B
B8
18
12
y
E
DB 10
I
L
AD
BE
A
E
D =
12
U
7
Q11
16
10
8
14
11
16
F
14
14
A
A
C
x
y
21
18
nABC
10 with12altitudes BE and AD
9
x
Prove:
R
HB
A
G
x
B
C D
C
8 S x
y
18
F
E 7 18F H
F x
T
Q
J
16
18
I
8
13.
D
M
A
|| BC
15. Given:
I
x
21 y
D
D
S
20°
N
15
R
16
3
I 10
T
y
C
C
x
7
8
J
A
D
C
D
Unit IV — Triangles
Part D — Similarity – Part 2 (Triangles and Their Parts)
Lesson 2 — Theorem 28: “If a line is parallel to
one side of a triangle and
intersects the other two sides in
different points, then it divides the
two sides proportionally.”
Objective:
To understand this theorem as an application of previously accepted
postulates, definitions and properties, and to demonstrate its proof directly.
Important Terms:
Auxiliary Element – From a Latin word meaning “to help”, this is a geometric
element which is not specifically referred to in a given diagram, but which
does exist, and is needed, in order to logically complete the demonstration of
a conditional. This element is usually drawn in a figure, using dashed lines,
and must be referenced in the proof with a step justifying its existence. It is
important to note that you must not place too many conditions on the
element (called “over-determining”) thereby denying its existence. Neither
must you place too few conditions on the element (called “underdetermining”), thereby allowing the existence of more than one such element.
In other words, a geometric element is considered to be “determined”, if
exactly one such element can be drawn to meet the conditions.
“Divides Proportionally” – Noting the illustration below, AB and CD are
said to be “divided proportionally” by points X on AB, and Y on CD, if
A
X
B
C
Y
AX CY
.
=
XB YD
D
A
D
Corollary 28a – “If a line intersects two sides of a triangle in different
M proportionally,
14
points
so that the sides are divided
then the line is parallel to B
2x
5
.75
1
9
3
the third side.” (This is the converse of Theorem 28)7
M
2
N
M
N
Q
4 three or more parallel
N
12 – “If
Corollary 28b
lines 65intersect two
transversals,
3
1
then they divide the transversals proportionally.”
R
M
3
3
N
N
M
2.5
1
N
Part D – Similarity
- Part 2 (Triangles
and Their Parts)
5
2
4.5
5
6
10
M
2
3
M
351
12
2x
dA
W
X
8
A
X
B
A5
2X
C
A
B
A9
B
Q
X
C
8
B
C
M
16
5
C
X10
A
B
16
2
Y
X
CY
Y
Y
5
10
BD
Y
D
4 4
A
X
B
A8
C
X5
B
C
5
Y
Q
D
Y
D10
10
16
15. DE =89, AB =5 12, BC = 16, EF = _____
A
CB
3
10
D
A
D
E
A G
4
D
B
3
E
Y
D
G C
10
F
D
X R BY
C S D
Y
Y
X
D16
A
7
4
10
14
S
8
5
16. DE = 9, EF =
BC = 40-x
F
7
4
10N
14
M
C
C
Q
NB
C
2 Y
17. AC = 16,QAB = M
4, DE A= 6, DF =X_____
A
D
X
7 1
4
10
14
A
E D
1 BB
A
B 4
18.
AB = 4, BC
9
G E
R
S = 5, FG
B = 6, ED =Q 8, EF = _____
X
A
A
D
2
4
G
R
S
1
F
X C
4
E
2
B
X
Y
C
N
F
4
3
3
A
B
1
7
X
A
G
Y
N
R
C
M
Y
S
X
Y
3
3
3
A
C
C
F
2.5
For Exercises 19-23, apply Corollary
28a
to
the
given
triangles
to
determine
if
XY
BC.
Y
3
1
B
2
N
M
A
4
C B
B
B
C
A
1
2
A
2
4X
19.
20.
21.B
2
C
C
3
3
X 4
X
2
2
X
Y
B
A
X
4
X
Y
3
3
A
A
X
4
Y
A
C
9
3
3Y
X
A
3
2
A
9
C
10
9
B
Y
4
3
6
8.4
X
1
Y
2
3
6
C1 A
B3 B 1
8.4
2A 7
C
C
C
B
Y
C
2
Y
C
B
Y
X
C
3
A
B 1 Y
7
X 2
3.5
2.5
Y
X
1
2
X
D
C
C
2.5
A
B
C 3.5
B
E
C P
C
9
A
A
2
22.
23.
B
2
X
9
C
10
X A
3
3
6
CA
R
C
B
11, AB = x,
B
X
A
||
8.4
3
X B
B
9
A
Y
1
B
Y
BB
B
Y
10
B3
In a9 new subdivision,
there is a section
of lots of X
10
B
X
Yruns along a
P
equal width, where the frontX lot-line
C
Y
C
9
north-south
street,
the
side
lot-lines are
10 all
C
Y
B
C
S
T
ek
3
X
D
69’
138’
G
Y
69’
X Surry Road
Surry Road
45’
135’
45’
X
144’
Surry Court
Z
135’
69’
Y
Part D – Similarity - Part 2 (Triangles and 138’
Their Parts)
Surry Road
45’
Oak St.
X
Y
45’
45’
138’
Y
135’
Oak St.
144’
Surry Court
Z
Oak St.
Surry Court Y 144’
135’
Z
Z
C
Bell Ave.
Surry Road
W
Oak St.
69’
138’ Surry Road
Z
Oak St.
Y
69’
3.5
Z
C
B
Bell Ave.
Bell Ave.
Bell Ave.
138’
Surry Court
urt
Surry Co135’
Z
Z
144’
Y
F
Y
A
Bell Ave.
perpendicular to the street, and the
runs
D back lot-line
Z
Z
along a Ccreek that
the lots atD aN northeast
D crosses
C
A
MB
R
angle as shown in the diagram. Note: WZ = 372 ft.
What is the approximate length (to the nearest foot)
of each rear lot-line?
144’
C
Y
D
W B
W
D
X
8.4
X
B
XC
C2.5
A
A
B h
C
C
C
A
3
6
3.5
A
A
B
Bell Ave.
3
B
Q
W
3.5 Y
ek
B
2.510
7
10
Cre
Y
X
24.
9
A
3
X
W
A
Y
4
9
A A
ek
A
YX
2.5
1
2
X
A
7
Y
14
8.4
9
ek
1
M
Cre
B
A
7
Y
Y
Cre
3
1
X 6N
3
ek
B
3
Cre
X
Cre
M
10 1
4
1
D
D
14
C
10
C
Y
For Exercises
14-18,
complete each statement
using the
following
diagram.
B
5
2
4
7
4 10
14
10
4
14. AB = 3, BC = 6, DE =A 2, EF = _____
Q C
X
B
Y X
7
D
D
N
16
X
355
X
Important Terms: (continued)
Geometric Mean in a Proportion – In a proportion, when the second and third
terms are equal in value, that value is called the geometric mean between the
first and fourth terms.
Corollary 30a – “If you have a right triangle, then either leg is the geometric
mean between the hypotenuse of the triangle, and the projection of that leg on
that hypotenuse.”
Corollary 30b – “If you have a right triangle, then the altitude drawn to the
hypotenuse, divides that hypotenuse into two segments, in such a way that, the
altitude is the geometric mean between the two segments formed by drawing
that altitude.”
Corollary 30c – “If you have an altitude drawn to the hypotenuse of a right
triangle, then the product of the lengths of that altitude and the hypotenuse, is
equal to the product of the lengths of the two legs.”
Example 1: Simplify each radical.
a)
Solution:
a)
17
2
b)
32
c)
32 = 16!•2 = 16 • 2 = 4 2
or
32 = 22•22•2
2•22 = 22 •222 •2 = 22 • 22 • 2
2•2
= 22•2
2• 2
=4 2
b)
2
17
17 • 2
17 •2
=
=
=
2
2
2• 2
2
c)
2
5
364
=
2• 5
5• 5
=
2 5
25
Unit IV – Triangles
=
2 5
5
34
2
2
5
Unit IV — Triangles
Part D — Similarity – Part 2 (Triangles and Their Parts)
Lesson 5 — Theorem 31: “If you have a given
right triangle, then the square of
the measure of the hypotenuse is
equal to the sum of the squares
of the measures of the two legs.”
(The Pythagorean Theorem)
Objective:
To understand this theorem as an application of previously accepted
postulates, corollaries and properties, and to demonstrate its proof directly.
Important Terms:
Corollary 31a – “If you have a right triangle whose acute angles have measures
of 30° and 60°, then the measure of the hypotenuse is twice the measure of the
shorter leg, and the measure of the longer leg is 3 times the shorter leg.”
(This special right triangle is usually referred to as a “30-60-90 Right Triangle”.)
Corollary 31b – “If you have a right triangle whose acute angles each have
measures of 45°, then the measure of the hypotenuse is 2 times the measure
of either leg.” (This special right triangle is usually referred to as a “45-45-90
Right Triangle”.)
Example 1: If a 25 foot ladder, represented by AB, is placed 7 feet, represented by AC,
from a vertical wall, represented by BC, how high up the wall will the ladder
reach?
B
Solution:
a2 + b2 = c2
b=7
cC = 25
a)
A
a2 + 72 = 252
25
a2 +449 = 625
2x
a = 576
C
7
A
A
a = 24x
5
h
2
So, ladder6 –will
reach 24 feet up
p the wall.
x
B
H
6
X
Part D – Similarity - Part 2 (Triangles and Their Parts)
x
z
369
2525
25
252x
4 44
4
2x 2x
2x
2x2x
4
5 55
5
h hh
h
x Ax
7
xx x
A AA
C CC 7 77 A AA
A
H HH
A argument explaining
Write aC paragraph
7
H
4.
6–x
x
A
C
h
6 – x6 – x
66 ––6xx– x
why the
HB B
45-45-90B Btriangle
6
B
corollary must be true.
6 66
6
5. Using the figure to the right, find each missing length. Simplify all radical expressions.
a) If x = 6 and y = 8, then z = _____.
X
X XX
b) If z = 15 andX x = 9, then y = _____.
c) If z = 15 and x = 10 , then y = _____.
d) If y = 2 and x = 45°, then z = _____.
x xx 30°
3
Z
45°
xx
30°
45°
30°
30°
45°
45°
30°
Z ZIfZ x 45°
30° Y W
e)
Z = 2 3Wand z = 6, then yYY=Y _____.
W
WW
z
x
z
Y zzz
y
y yy
y
6. In the diagrams
below, find x.
A
x
4 x xxx
14
A AA
A
1414
14
14
a)
2x 2x
2x
2x2x
4 44
4
b)
2x
x
12
x xx 12
12
x 12
12
c)
x
x xx
x
6
6 66
6
12
24
2424
24
24
1212
12
12
8
d)
8 88
8
e)
12
x xx
x xx
x
x 10
10
10
10
1212
12
12
x
x
10
x
x
x xx
x
x xx
x
45°
30°triangle?
7. Which
45° 45°of these triples could be sides of a right
45°
45°45°
z zz
z
x
x x
x 3, 4)
a) x(2,
y yy
y
b)
45° 45°
45°
45°45°
(
30° 30°
z
30°
30°30°
2 , 3 q, 5
q
n nn
n 45° q q
)
n
1
q 2)
c) (1, 1,
1
1
d)  , , 
 3 4 5
y
60°
60° 60°
60°
60°60°
m
m mm
m
B
B BB
B
C
A AA
A
6
C CC
C
A
12
1212
12
12
D30°
660°66
6
60° 60°
60°
60°60°
A
A AA
A
D
D DD
D
D DD
D
30°
30°
30°
30°
30°
C CC
C
C
B
C CC
C
a
h
b bb
b
h hh
Unit
IV
–
Triangles
h
m
n Bnn
m mm
n
m
A AA D
B BB
B
b
a aa
a
372
B BB
B
B CBB
B
n
X XX
X
Y
Y YY
Y
X
P
PAPP
P
A
C
X
25
4
2x
12
5
x
10
h
y
A C
45°
Refer to the figure at the right to find the missing14
lengths in Exercise
8.
Z H
x
a)
b)
c)
6
_____
2.5
z2x
y
_____
10
x
4
2
_____
Z
e)
_____
_____
7 2
_____
_____
2x
y
12x
x
30°
W
m
12
x
24
B
8
10 C
a)
4
b)
2
3
xq
2x
_____
_____
x
12
12
_____
12
z
C
_____
d)
_____
_____
3
4
e)
_____
5 3
_____
f)
_____
9
6
m
45°
B
n
x
60°
26
C
T
D
X
Y
C
A
A
a
6
25
C B
y
x
D 26
D
C
30°
10
C
D
T
B
Z
b
25
12
Y
26
X y Parts)
Part D – Similarity -hPart 2 (Triangles and Their
B
m
n
A
B
6
A
A
17 –x
a
X
R
12
n
W
{
A
C
b
h
m
B
D
60°
C
P
60°
>
B
12
D
B
A
For Exercise 11, find the indicated measures,
referring
to the m
figure
at the right, where
D
C A
Z
DB.
m/BAD = 105° and AC
AC = _____
BC = _____
DC = _____
AD = _____
m/BAC = _____
m/CDA = _____
B10
A
CB
6
D
A
n
m
17 –x
60°
q
h
m
30°
C
x
B
b
q
n
A
B
30°
n
8
30°
12
D
D
2510.
Rectangle ABCD is shown to the right.AUse this figure for Exercise
y
y
A
x
60°
45°
10. a) Find the length of the diagonal
AC.
C
z
b) Find the length of side BC
x of the rectangle.
c) Find the perimeter of rectangle ABCD.
X
y
6
a
45°
x
A
10
A 45°
C
x
10
b
60°
x
_____ B
C
x
mzD
x
12
D
x
h
y
_____
D
q
n 12
a
45°
x
6
30°
24
6
x
x
45°
x
_____
c)
11. a)
b)
c)
d)
e)
f)
A
y
A
n
60°
Y
x
m
x
z
A right to find the missing lengths in Exercise
60°
Refer to the figure at the
9. 4
14
9.
q
n
12
45°
45°
_____
f)
14
24
_____
d)
30°
z
x
A
_____
_____
6
45°
X
_____
Y
x
12
x x
x
B
30°
W
6
8.
x
6–x
x
A
A
7
A
x
P
17 –x
373
10
E 45°
Cube
A
F
F
B
8
2x
8
x
B
17
45°
D
15
ECube
C
A
x
A
2x
H
B
P
C
h 15
45°
Cube
X
6
D
Y
Q
A
A
2x
P
X
10
C
2x
Q
M
W
6
P
A
A 5.
6
3 Y
16
45°
6
Z
A
N
2
A
1
Y
4
Q
Find xPand y.
2x
B
h
A
2x
6
R M
6
W
A
P
4
2x
V
P
A
2x
E
Q
y
45°
x
x
Y
N
N
T6
Q
A
2x
x
Q
A
C
G
UH
Note: This is a right
M
D 8x
NC
rectangular
prism.
45°
6
Cube
y
N
T
M R
4
P
16
Q
4
QB
x
V
A
378
E
6
D
C
x
60°
A
F
H
G
C
h
X
Y R 1 X
Y
A
Q
X
E
F 61
N
6
1 x M
6
W
Z
B
h
6
4
x 1645°
N
1
H
G
12
Q
Q
B1
3 T
1
P
A
P
M
1
Q
x
10
1 R Y
1
2x
A
6. FindXthe perimeter of the
base. N
1
M
1
Find altitude EQ.
B4
E
h F E16
6
x
P
N
Q 1
B
A
P
3 S T 8e
1
T 8
8
8
M
B
1
Q
8 G N N
M
D
M
A
Y RG1
1
H
6
A
C
X
Q
e
y
T1
R
60°
2
A
1
2x E
F
B
x
e
D
C P
4
2Q
F
N
x
1
x
D
B
A1
V
U
e
P
y
P2 (x2 , y2)
M
1
Q
dH
U
17
Note: This is a right
8 prism. N
M rectangular
15
6
D
y
T
R
Unit
C IV – Triangles
P
Q
4
6
14
Q
A
1
B
h
X
1
N
1
M
1
Q1
4
B
A Z16
F
10
2x
2x
8
10 MX
10
10 1J
X
xD
G
P
12
x. I T
3
P
10
15 h
h D
x
1
26
F
triangle.
2x
2
17
x
C
2
R
x
x
6
C
45° P
F
4.Q Find
Z B
W
Cube
J
C
x
A
H
6
3. Find altitude
h and slant height ,.
6
A
B
A
E
45°
8
8I 17
G
15
12
Q
10
A
C
G
D
G
D
C
I
8
H
8
17
2x
2.
E
6 F
Q
60°
J
2
x
Show Ethat nABC isF not a right
2
F
x 6
x
B
D
A
B
A
B
Q
C 10
D
G
H 8
8
G
2x
H
Lesson 6 — Exercises:
A
B
D
A
1. Find x.
8
2 x
x
10
AA
y
H
A
H
D
e
2
e
x J
C F
R1 (x2 , y2)
x
D E x2e–E x1
P1 (x1 , y1)
6
Note: This isI a right
rectangular
G prism.
y
E
By
P2 (x2 ,e y2)
12
6
F
y
d
G
D 15
Q D
A
C
B
C
x
I
D
2x
10
A
A
X
J
12
C
xW
6
6
2x
E
Y6
G D
F 6
A
6
x
Q
H
x
V 3
1
12
Q
F
6
T
G
G
P
Y
1
h
R
X
T
M3
1
h
1
X
10 x
2x
6
B
N
N
Y
1
R
1
x
h
A
6
x
1
h
Q
Use the figure below to find
4 asked for in1Exercise 9-14.
6 lengths
h each
16 of the
Y
6
6
Q X
1
1
Z
x
A
AN
N
1
4
x
2x
x
6
Y
3 T
W
1
1
4 9. NQ
N
Q
ME
y
1
Q
11. YQ U
8
M
B
QN
A
x
T
E
1
P
N
12. QAQ
P
4 B
H
A
Q
U
e
y
1
e
1
P
M H 1 14. PQ G
13. BQ
Q
e
D
C
F
E
P2 (x2 , y2)
B
y
A
d
ex
U
y2
d
1
e
Q
G
Note: This is a right
rectangular
prism.
e
P2 (x2 , y2)
C
H
y
D
e
A
B
B
N
G
VC
e
D
P
1
F
e
A A each edge of the
15. Suppose
e cube at the right
6
E
a Rformula Ffor
of
is e. Develop
y the measure
T
the distance
from point A to G in terms of e.
2x
UT
Y
X
N
A
14
F is a right
Note: This
Q
rectangular prism.
1
T
1
8
x 16
1 10. XQV
3
Q
1
1
P
T
1
B
N 1
B
N
M
Q
1
P
M
1 4 Q
M
N
Q
Z
A
1
45°
P M
Y
1
6
X 10
R
2x
Q
right
rism.
10
P
Y
right
sm.
R
J
Z
12 R
A
A
G
I
45°
H
D
12
F
15
Q
6
2
x
8. Find h. Find x.
B
6
17
h
x
H
E
6
D
P
I
E
6
F
H
D
J
10
A
2
x
7. Find x.
A Find m/JHG.
B
6
x x
60°
45°2
Cube
F
6
DCC
y2
G
x
H
x2 – x1
D
e
R1 (x2 , y2)
C
y
P1 (x1 , y1)
e
y
E
P2 (x2 , y2)
D (3, 4, 0)
P2 (x2 , yx2)
d
x2 – x1
P1 (x1 , y1)
d
R1 (x2 , y2)
y2
F
C (0,00, 0)
x2 – x1
y2 A
x
x
y
x
P1 (x1 , y1)
R1 (x2 , y2)
x2 – x1 Part D – Similarity - Part 2 (TrianglesGand Their Parts)
B (3, 0, 8)
R1 (x2 , y2)
P1 (x1 , y1)
y
D (3, 4, 0)
E
379
16
Q
M
1
Extending Your Mind:
8
M
A
Q
F P2 (x2 , y2)
E y
N
6
Recall from Algebra (Videotext
Algebra: A Complete Course,
y
T
R
Unit VII, Part D,2x Lesson 1), the mathematical relation which
A
is obtained by applying4 theP Pythagorean Theorem
to two
Q
x system.
points in the Cartesian coordinate
V
The Distance Formula:
2
(
(
d = x2 − x1
U
) + (y
) (
2
2
− y1
x
G R1 (x2 , y1)
P1 (x1 , y1)
H
)
)
e
x2 – x1
2
Note: This is a right
2
rectangular
d = x 2 − prism.
x1 + y 2 − y1
y2 – y1
B d
e
2
e
D
y
(
C
P2 (x2 , y2)
)
(
)
The distance between two points in three dimensions, P1 x1, y1, z1 and P2 x2 , y2 , z2 , can be
found using a formula similar to the distance formula.
d
y2
d=
(x
2
− x1
) + (y
2
2
− y1
)
2
+ ( z2 − z1 )2
x
x2 – x1
P1 (x1 , y1)
R1 (x2 , y2)
y
Consider the rectangular prism to the right, sketched in a
three-dimensional (x, y, z) coordinate system
for Exercises 16-19.
16. Write the coordinates of vertices A, C, E, F,
and G.
A
F
G
17. Find the length of diagonals DF, DG, and DB.
z
18. Give the dimensions of the prism.
19. Find the volume of the prism.
380
Unit IV – Triangles
D (3, 4, 0)
E
C
0
(0, 0, 0)
B (3, 0, 8)
x
20. Five of the eight vertices of a rectangular prism are A (1, -1, 2), B (1, -1, 6),
C (1, 5, 2), D (1, 5, 6) and E (9, -1, 2).
Find:
a) The dimensions of the prism.
b) The coordinates of the other three vertices.
c) The length of diagonal DE
Note: It might be helpful to sketch the prism in a three-dimensional coordinate
system.
Part D – Similarity - Part 2 (Triangles and Their Parts)
381
G
R
S
U
D
T
V
W
E
F
Lesson 1 — Exercises:
S
T
A
B
V C
W
G
1. In the figures to the right, nMNQ > nPRT.2x Complete each
D statement below by
supplying the missing symbols.
E
G
P
M
a) The correspondence M __ __ to __ RT
F
D
is a congruence.
A
C
B E
b) The correspondence QM__ to __ __ R
F
2x
is a congruence.
A
C
c) /M > /___
B
A
P
M
d) /Q > /___
R
N
Q
2x
e) /N > /___
2x
f) MQ > ___
M
P
g) NQ > ___
T
h) NM > ___
A
XQ
i) nTRP > n____
R
ZN
T
T
Y
2. In the given figure, nTXY > nSXZ.A Use this correspondence for a-f below.
N S
R
Q
a) /YXT corresponds to /__ X __.
b) /YTX corresponds to /__ __ X. 2xA
X
Z
c) /TYX corresponds to /__ __ __.
Y
M
2x
T
d) Is a line segment congruent to itself? Why?
e) Is an angle congruent to itself? Why?
S
X
Z
B
C
KY
f) Is a triangle congruent to itself? Why?
2x
T
T
Y
H
A
S
3. Given nBCY > nHKM. Determine which of the correspondences
below are
A
YB
M
2x
A
congruences.
a) /BYC corresponds to /HMK.
B
K
YC M
b) /BYC corresponds to /HKM.
2x
c) /YCB corresponds to /HKM.
D
C
d) /YCB corresponds to /MKH.
A
BK
B
C
e) /CBY corresponds to /KMH.
f) /YBC corresponds to /MHK.
A
B
g) /CBY corresponds to /KHM.
h) /BCY corresponds to /HKM.
D
C
D
H
H
C
Part E – Congruence - Part 1
385
Y
2x
M
B
B
C
H
K
4. In the figure below, nABC > nCDA. List the congruent parts.
A
B
A
D
D
C
5. In the diagram to the right, nABC > nXYZ. Use this relationship to complete each
of the following statements.
Z
CC
C
E
D
P
a) XY = 2____
X
10
30°
B
b) m/Y
= ____ U
X
14
N
M
c) m/C
= ____
3
A
N
4
d) ZY = ____
75°
1
2
T B
Q
A
8
h
B
A
Y
B
X
Y
A
e) m/A = ____
F
f) m/B = ___
2x
H
S
B
R
U
E
6. Use the figure below to name three
pairs of triangles that appear to be congruent.
P
C
TE
S
S
T
A
N
V
C
14
D
M
W
G
N
R
H
4
F
Q
GA
h
H
B
D
S
F
S
E
G
F
A
B
T
C
2x
G
N
2x
7. Use the figure and your answers for Exercise 6 to complete each of the following
M
P
X
Y
Z
statements.
a) /FES > ____
b) FE > ____
c) /HFE > ____
d) GS > ____
W
N e) /EGS >
Q ____R
T
f) FS > ____
C
T
Z
386
X
Unit
IV – Triangles
Y
A
1
D
2
B
2x
B
A
V
X
S
T
U R
Q
T
R
F
M
C V
UP
W
F
N
A
R
Q
R
D
P
C
4. Prove
Postulate Corollary 13a –“If W
two angles and a non-included sideT ofXEone
M R
Y
F
triangle are congruent to the two corresponding angles and the corresponding
nonE
Q
T
included side of another
triangle, thenV theT twoUtriangles are congruent.” (Note: This is C
S
R
MD
R
B
B P it as Nan exercise
D
the same postulate corollaryAwe proved
in Ethe lesson.
We are using
S
M
Q
Y
Z
R you understand
X its proof.
Y your course notes to check.)
to make Ysure
Use
W
V
B
U
T
Q
A
C
a) State the corollary.
Q
A
A
B
FN
E
b) Draw
and
label
a
diagram
to
accurately
B
T showXthe conditions of theA corollary.
W
2x
M
N
Y
c)C List the given information.
P
D
A C
d) Write the statement we wishQ toQprove.
C
F
F
U
V V D TW U X
F
R
R G using the two-column
M
e) Demonstrate the direct proof
format.
NF
P
N
A
Q
D
F
EA
AD E
F
M
C Y
D
Q is the
of FM
B midpoint
Z
Y
U
C
E
X
W
V
W D
5. Name the
pairs
of
triangles
that
are
congruent
and
name the Postulate
E13 Assumption
T
E
Q
A
Q
B
A R
N
or Corollary
which justifies the congruence. S
x
B
AW
B
BC
E
D
B
A
T
G
Q
A X
H
G
J
K AS
C
R
P
2x
M
A
Q
D N
D
Y
a)
b)
c)
A
C
X
R
CH
F
A
H
J
G C
Q
2x
A
F TD
E
F
A
RG
E E
YBF
C M
T
I
L
D
S
Q is the midpoint of FM
M
YA
B
D
Q
R
UW
U V R VT
B
C
X
R
M
I
S B
P B
NQ
D
A
E P
DQ
R
M
F
I
J
C
B
D
H
A B
C
T
G
W
Y
M H
V
G
F
U
T
E P
A
P
C
H
Q
A
D
T
A
P
A N QF
A
R
E
D
d)W
e)
f)
B
D
A
X
R
K
J
H
H
G
J
K
F 2x
D
Q
C
O
N DR
J
F
2x
U
V GT
F
C E
DR
O
C
M
F
F
A
M
A N R
E
U
N
P
G
X
V
W
J
S
I
T
E
R
S
R
E
M
2x
Y TC B
WRS
Q
For
nUWY
and
nXVY
D
E
C
B
A
T
B
E
Z
X
Y
S
G
Q
C
Q E
E A
B
B LB
AI D
N
B
where
UV >I XW
RQ C
A
I
A
A
I
J
C
X
S
A
B
B U
E
D
U
S
T
P
P
M
F
Q
X
H
XD T
G
J
K
C
x
M
H
N
R
C Y
F
A
D
R
D
C
G
D
A
CD
A
g)
h)
i)
MN
>
RP
C
F
V
V
Q
A
D
U
F
F
E E
R M
M
YN
E
U
A
V
U
U
P
D
C
R
JT
W
M
G T
FW
V
V
2x
N
CQ S B
D
Q
CA
O P
F
N R
RI
W
L
X
Q is the midpoint of FM
G
W
X
B
A
D
Y
U AS V W B W EEX Q D
D
Y V
B
Q
Z
T
U
E
B
R
A
Z
Q
B
N
A
A
P
C
M
T D
F
X
Y
C
Q
B
H
E
T
E
B W
G
X
Q
D
G
T
I
H
AA
F
E
C
F
A
B
C
A
S C
AZ
F
U F
Y
j)H
AE > DB
D
C KK
H
X
G
J
J
Y
Z
E
U
V
F
M Y
Q
O
T
D N RY M
R
U
D
M
C
A
X
P
N
V
G W
Z
X
Y
Q
V
J
I
Q
W
C
D
D
A
B
E
D
BU B
E
D
C B
DE
Q V B TW
A A B
DB
C
NR
E
I
L
A
W
A
2x
A
X
M
N XI
B
Y
I A J
392
Unit IV – Triangles Y
HU J Z
GS
F
E
K
C
Q
P E
BF
M A
A
CF
F
XD
A
C
H
A
Q
X
Unit IV — Triangles
Part E — Congruence – Part 1 (General Geometric Relationship)
Lesson 3 — Congruence Postulate Corollaries
Objective:
To understand the applications of the Congruence Postulate to the
relationships between right triangles, and to prove those applications directly,
as additional Congruence Postulate Corollaries.
Important Terms:
Congruence – From the Latin, meaning “to agree”, this is a relationship
between geometric figures which have the same shape, and the same size. For
practical purposes, two line segments are said to be congruent, if and only if,
their measures are equal. Additionally, two angles are said to be congruent, if
and only if, their measures are equal. Finally, two polygons are said to be
congruent, if and only if, for some pairing of their vertices, the corresponding
angles are congruent, and the corresponding sides are congruent.
Postulate Corollary 13b – “If the hypotenuse and one acute angle of one right
triangle are congruent to the hypotenuse and the corresponding acute angle of
another right triangle, then the two right triangles are congruent.” (This is often
called the “HA Postulate Corollary”.)
Postulate Corollary 13c – “If a leg and one acute angle of one right triangle are
congruent to the corresponding leg and acute angle of another right triangle,
then the two right triangles are congruent.” (This is often called the “LA
Postulate Corollary”.)
Postulate Corollary 13d – “If the two legs of a right triangle are congruent to
the two corresponding legs of another right triangle, then the two right triangles
are congruent.” (This is often called the “LL Postulate Corollary”.)
Postulate Corollary 13e – “If the hypotenuse and one leg of one right triangle
are congruent to the hypotenuse and the corresponding leg of another right
triangle, then the two right triangles are congruent.” (This is often called the “HL
Postulate Corollary”.)
Part E – Congruence - Part 1
395
A
F
A
M
T
V
T
V
M
X
2x
H
RX
V
U
G PR
Z
X
Example 1: Indicate in each of the following Awhether theDgiven pair of triangles is
U or LL Postulate Corollaries.
N
congruent by the HA, LA, HL,
C
T
B
a)
A
b)
U
H
2x
A
X
D
2x
B
M
Q
N
d)
X
U
2x
U
Q
Solution:
a) HL
b) HA A
R
G
Q
P
U
2x
T
V
P
N
V
MH
2x
T
R
M
A
P
c) C LL
P
Q
N
N
V
Q N
X
e)
LA
Dd) HL, RHA or LA Q
AP
A
U
G
B
QM
PY
I
M
Example 2: State the additional information
H you would
G need to
2x
A following
prove nABC > nABD, using
each of the
2x
A
B
corollaries.
A
Q
X
Y A
a)C HA
A
P D
A
D
2x
N
b) HL
c) LL
M
2x
d) LA
P
B
Solution:
2x
A
A
A
2x
C
BA
UE
F
D
T
a) A/CAB > /DAB
orAA/CBA >
/DBA
D
C
D
U
b) AC > AD or CB > DB Q
V
R
E
c) AC > AD and BC > 2xDB
2x
V A
X
d) AC > AD or CB > DB and
B
C/BAC
A > /ABD A
B E> /BAD
F
B A
or /ABC
C
A
U
C
X
D
D
D
W
Q
P
M
U H
D
A K
D D V C
CB
D
M
I
J
Y
H
Q
DA
C
X
D
M
C
2x
A
A
BW
J
B E
U
W
EC
A
D
F
B
N
MO
MA
G D
D
B
DN Y
P
MO
B E
IE
X
V
X
X
R
A C
D
F
UB E
D
H
U
K
C
G
A
P
Q
Q
M
A
D
A
B
A
C
MA Q
X
2x
D MQ
TX
BE
2x
2x
C
M
A
P
G
C
Q
P
BC
E
I
E
A
D C
A
F
B
C
D
V
DA
C J
V
F
O
A
B
C
W
BH
D
A
F
D
A
M
D
H
2x
D
X
C
A
B E
N
F
A
U
A
W
H
D
C
F
W
E
A
N
B
A
A
N
B
C
J
D
E
J
A
V
Unit IV – Triangles
B N
O
B
C
D
X
H
E
AJ
B
M
2x
396
V
E
E
J
C
D M
C
F
G
M
N N
M
T
C
M
F
e)
W
UD
D
M
B
H
R
X
PXU
R
Z
H M
V
T
V
G
F
A
N
c)
M
U
B
A
1
2
A
Y
Example 3: Supply the missing reasons in the following proof.
B
B
>
Given: BC
AD
DC > AC
P
S
Prove: nABC > nDBC
A
STATEMENT
1. BC
C
D
M
REASON
> AD
1. Given
2. /DCB is a right angle
/ACB is a right angle
2. _________________________
3. nDCB is a right triangle
nACB is a right triangle
3. _________________________
4. DC > AC
4. _________________________
5. BC > BC
5. _________________________
6. nABC > nDBC
6. _________________________
Solution:
2.
3.
4.
5.
6.
Definition of perpendicular lines
Definition of right triangle
Given
Reflexive Property for Congruence
LL Postulate Corollary
Lesson 3 — Exercises:
For Exercises 1-3, state the given postulate corollary, draw and label a diagram to
accurately show the conditions of the corollary, state the given and the prove, and write
the demonstration of the corollary.
1. Postulate Corollary 13-b: Hypotenuse - Acute Angle Postulate Corollary
2. Postulate Corollary 13-c: Leg - Acute Angle Postulate Corollary
Part E – Congruence - Part 1
397
R
M
2x
AB
N
P
2x
B
U
C
T T
B
A
3. Postulate Corollary 13d: Leg - Leg Postulate
Corollary
X
Q
V
A
A
C
D
Q
R
A
C
D
A
A
X
D
P
I
T
R
2x
Z
H
D
K
VD
Y
P
I
V
U
D
5.
D
C
A
D
B
F
A
2x
2x
C
B E A
U
R
F
D
A
T
2x
M AV
H
G
9. V
V
2x
A
X
R
N
M
BR
P
T
V
P
F
M
RN
A
N
E
V
P J
2x
A
F
C
398
X
A
2x
A
X
G
Unit IV – Triangles
A
D
M
Q
D
W
B
B
P
C Y
H
K
A
HD
D
2x
C
B E B DF
V
N
G
D
C
M
P
B
N
N
M
P
T
A
D
Q
T
M
R
Q
H
K
G
I
J
Y
C
D
O
B
A
C C C
B
C
B
N
M
VP
C
A
D
B
U
F
A
C
CM
F
H
Q
A
G
E
X
N
A
RE
B
Q
A
A
N
WD
B
M
U
G
N
H
RU
P2x B
RA
K
U
QN
T
D
M12. X
11.
Y
Q
B
C
A
P
C
V
D
UQ
I
N
J
M
A
V
N
G
P
Q
R
A
M
2x
A
Q
M
D
C
X
Y
A
P
C
D
A
B
H
G
P
A
G
H U D
T
N
2x
M
A
A
D
Q
M
X
2x additional information
For Exercises
13-16, state the
you would
need to
A K
P D
C
D
Q
QD
V
X
Y
B
C
of
the
indicated
postulate
corollaries.
prove nACD >PnBCE, using each
P
Q
B
A
O
R
M
I B
JE
N
D
M
2x
A
D
M
13. HA
C
A
B E
FD
B
CC A B
14. HL A
G
H
A
D
E
P
U
15. LA DU
A
D
K
A
T
N
M
Q
X
YC
D
16. ALL
2x
P
C
D
C
A
B E
F
Q
B J
I
V X
C
E
V Q
A
P
R
B
A
O U
E
M
T
P
M
D
X
U
B
R
M
H
F
G
M
X
C
2x
XM
10.
W B
U DQ
B
C
HR
T
J
M
N
X
A
C
D
HF
XA
BV
H
U
V
7. M
O
T
D M
H
F
D
A
C
J
Z W
X
D
A U
G
N
U
D
M
8. X
C
V
P
Z
W
X
N
U
B
H
EX
T
X
U
M
M
V
A
6.
B
O
T
V
TU
W
P
2x
F the R
For Exercises 5-12, name the pairs of triangles
that
are congruent.
Then
CX state
D
Z
A
D
M
C
U
A
E
B
F
B
W
X
N
R (HA, LA, HL, or LL) which justifies each congruence
X
W
B
postulate
corollary
relationship.
C
Z
A
D
C
C
A
V
E
A
X
J
EX
R
X
M
2x
D
UA
Q
G
4. Postulate Corollary 13e: Hypotenuse - Leg Postulate Corollary. This corollary is
TM
C
U in
included2xin this lesson because
it is part of Bthis family
of
postulates
2x
W
X and corollaries
R
X
D
Z
A
D
C
B
E
F
A
Lessons 2 and 3 aboutA triangle congruence.
State
the
corollary
and
sketch
a
diagram
G
A
H
B
to accurately illustrate the conditions necessary.
A demonstration
or proof
U Tof this
C
B
KD
A
Q
A
D
F
T will be asked
X for in
corollary
Unit IV,PPart
F, Lesson
4.
Y
M
V
2xU
V
D
M
C
C
A
A
17. Given:
Prove:
18. Given:
Prove:
C
C
19. Given:
A
A
2x
2x
B E
B E
2x
2x > /CBA are right angles
/ADC
V
X
V
DC || BA X
C
C
A
A
B
B
nADC > nCBA W
W
D
D
H
H
M
ABCD is a M
parallelogram
AC; BF
AC; AE > CF
DE
>
J
J
nAFB > nCED
N
N
T
T
B
CB
B
B
A
A
D
D
>
PR
MN
V
V T
V of MN
R is midpoint
V
Prove:
nMRP > nNRP
U
U
21. Given:
>
UV
U
U
U
Prove:
2x
22. Given:
2x
2x
2x
Prove:
23.
24.
C
C
2x
2x
A
A
F
F
X
X
F
F
F
F
X
X
X
X
>
U
U
B
M B
M B
M B
M
PNTT
T
P
P
PP
PP
R
RR
RNNR
A
A
A
A
A
A
N
N
N
>
M
MM
MRMM
NNN
N
G
N
NPP
NQ
QM; QN N P NP; UUMQ >
U
QP
Q
U VV
Q
nMQN > nPNQ
V
Q
P
V
P
2x
X
X
AD
plane of
M)
nBDC is anHequilateralG triangle
nADB > nADC
WW
WW
W
W
XX
XX
M
M
H
G
M
H
G
M
nBDC
(Plane
MH
G
G
MH
H
V
VV
V
V
V
U
U
U
R
T
T
T
T
T
B
B
U
UU
Z
Z
X
E
E
A
A
T
T
T
T
UVWX isBB a square
C
C
F
F
>
V
A
A
D
D
> nWXV
a) 2xUsing the LA Postulate
Corollary
R R
X
X
Z
Z
A
2x
AD
D
R
X
ZR
A
D
X
b)
Using
the
LL
Postulate
Corollary
ZX
2x
A
D
R
20. Given:
B
B
C
C
U
U
C
C
A
A
C
A
Prove:
nUVX
A
2x
2x
A
A
F
F
T
T
R
R
R
R
B
MB
MM
M
M
MQ
Q
Q
Q
M
D
D
D
D
D
D
C
C
C
C
C
C
N
N
N
NP
P
P
P
N
A
T
G
N
H
M
A
G
H
Q
Given: m/GKH
=
90
V
A
K H
G
A
Q
P
K H
A
V P A at K
X each
G Q
Q other
HI Qbisect
RYY
A
C GJA and
D
X
K
P
Q
A
P
C
I
Q
J
D
R Y
X
K
A
I
J
Q
A
P
C
D
X
Y
I
J
A
P
C nGKH D> nJKI
M
Prove:
2x
I
M
J
D
C
2x
MA
D
C
G
H
2x
B
M
A
B
2x
GD
H CC
D
D
K
A
D
B
A
D
Q
Given:
Point
O
A /AB andX
D/B are right angles.
Y
K
A
Q
PDD
DA is the midpoint
D
>
OC
of
AB.
OD
X
Y
I
J
B
A
O
PE
D
D A
B
A
E
IO
J
B
A
O
E
B
A
Prove:C nDAO
> nCBOMA
O
E
B E
F
B
MA
D
C
C
C
B E
F
B
A
C
A
C
A
D
C
E
B
F
B
U
BA
C
C
A
BU E
F
B
C
BA
DD
U
C
2x
C
D
U
DD
2x
V
XD
C
D
2x
V
X
C Part E – Congruence D
2x
V
X
B
A
O
V
X
B
AE
B
A
W
B
O
A
E
W
B
Part 1
399
A
Example 1: Given: AB > AD
/1 > /2
A
2
1
Prove: BE > DE
Z
Write an analysis and demonstration of the proof. 2x
Solution:
W
D
B
C
E
A B a
C pair
Analysis: I can prove BE > DE if I can find
of congruent triangles that
D
contain these segments, as corresponding parts. nABE and nADE are two
C
D angles between
such triangles. Finally, since angles 1 and 2 are included
two congruent sides, we can use the SAS Congruence Assumption to prove
O
the two triangles congruent, and then use the definition of congruent triangles.
STATEMENT
W
Z
REASON
1. AB > AD
1. Given
2. /1 > /2
2. Given
B
A
B
A
3. AE > AE
3. Reflexive Property
A
1
2
W
4. nABE > nADE
4. SAS Congruence Assumption
5. BE > DE
5. CPCTC B
D
A
2x
Z
D
C
E
C
A B C
D
W
A
O
Prove: /DAC > /BCA
2x
Write an analysis and demonstration of the proof.
Solution:
C
D
Example 2: Given: AC and BD bisect each other
Z
D
A
C
B
B
A
Analysis: I can prove /DAC > /BCA if I can find a pair of congruent
A B C D
triangles that contain these angles, as corresponding
parts.B nDAO and
D
nBCO are two such triangles. And, since AC and BD bisect each other, I
A
have two pairs of congruent segments which are in nDAO and nBCO.
C
Then, since /DOA and /BOC are congruent vertical angles, I can prove
A
C
nDOA > nBOC by the SAS Congruence Assumption,
and DthenEuse the
B
B
A
definition of congruent triangles. (Note: nDAC and nBCA could also be used.) D
A
STATEMENT
2x
1. AC and BD bisect each other
2. AO > CO; DO > BO
2x
REASONC
D
C
B
G
A
1. Given
A B 2.
C D
A segment bisector divides a
F
E
H
segment into two
congruent
A
parts
B
A
5
6
1 2
C
3. /DOA > /BOC
7
E
3.
the
A B IfCtwoB lines intersect, then
D
A
D
B formed are
D
vertical angles
A
congruent
C
D
E
Q
2x
4. nDOA > nBOC
3 4
4. SAS Congruence Assumption
5. /DAC > /BCA
5. CPCTC F
E
Part F –
B
Congruence
A
H
B
C
A B C D
A
G
2x
-
QA
D
G
F
P
E
H
PartD
C
2
B
405
R
D
A
A
A
A
5. Given:
DC bisects /ACB
AC > BC
2x
2x
Prove:
AD > BD
A B C
D
A B C
D
>
DB
AC
DB bisects AC
B
O
A
A
/ABD > /CBD
Prove:
8. Given:
Prove:
BD bisects /ABC
BD bisects /CDA
/BAD > /BCD
>
AC
BD
DB bisects AC
AD > CD
D
D
C
C
A B C D
A B C D 2x
A
A
E
E
D
D
A
A
C
B
C
E
B A
HE G
H G
E
B
Q
Q
Prove:
/BDC > /EHG
F
F
3
Q
Q
B
B
D
D
B
5
6
1 2
7
D
1
2
1
4
E
A
AA
A
D
D
C
D
R
R
P
P
C
C
2
G
S4 3
SF
E
BC > EG; DB > HE
/FEG is supplementary to /DBC
C
B
2x
A B C
D 2x
3 E4
B
P
P
H
H
D B
B
3 4
C
C
C
A
A
C
G
DG
F
F
2x
9. Given:
7
A 7
D
D
D
B
2x
5B 6
5 1 26
1 2
E
E
B
B
A
A
B
A
7. Given:
A
W A
C
C
C
A
A
A
B
B
C
E
2x
Prove:
W
Z
Use the diagram to the right for Exercises 6-8.
6. Given:
C
C
Z
A
A B
D
D
D
B
B
2x
A B C D
D
D
D
2
1
C
C
P
HP
A
3A 4
E
Q
P
C
A B C D
B
B
D
A
D
A
2x
B
C
D
G
F
A
E
H
A
Q
1
R
P
2
Part F – Congruence - Part 2
4
S
3
P
407 C
Unit IV — Triangles
Part F — Congruence – Part 2 (Applications)
Lesson 3 — Theorem 32: “If two given
triangles are both congruent to a
third triangle, then the two given
triangles are congruent to each
other.”
Objective:
To understand this theorem as an application of the definition of congruence,
and to demonstrate its proof directly.
Important Terms:
Congruence – From the Latin, meaning “to agree”, this is a relationship
between geometric figures which have the same shape, and the same size. For
practical purposes, two line segments are said to be congruent, if and only if,
their measures are equal. Additionally, two angles are said to be congruent, if
and only if, their measures are equal. Finally, two polygons are said to be
congruent, if and only if, for some pairing of their vertices, the corresponding
angles are congruent, and the corresponding sides are congruent.
Lesson 3 — Exercises:
1. Prove Theorem 32 – “If two given triangles are both congruent to a third triangle, then
the two given triangles are congruent to each other.” (Note: This is the same
theorem we proved in the lesson. We are using it as an exercise to make sure you
understand its proof. Use your course notes to check.)
a) State the theorem.
b) Draw and label a diagram to accurately show the conditions of the theorem.
c) List the given information.
d) Write the statement we wish to prove.
e) Demonstrate the direct proof using the two-column format.
G
2. Given:
Prove:
414
nAGC > nFHD
AB > EC, GC || BI
AG || EI
nBIE > nDHF
Unit IV – Triangles
H
A
A
C
D
B
E
2x
I
F
4. Write a demonstration (proof) for Postulate-Corollary 13e, the HL Congruence
Postulate (Unit IV, Part E, Lesson 3) by supplying the missing reasons in the following
outline.
Postulate Corollary 13e - If the hypotenuse and one leg of one right triangle are
congruent to the hypotenuse and corresponding leg of another right triangle, then the
two right triangles are congruent. (HL Congruence Postulate)
I
F
A
H
G
E
D
2x
nDEF is a right triangle
with right angle Iat E
F
F right angle at H
is a right triangle with
I
DF > GI; EF > HI
Given:
A B CnGHI
D
A
H
G
E
E
D
D
Prove: 2x nDEF > nGHI
H
G
A B C D
F
A
I
B
E
D
H
G
F
D
B A
A
C
15°
E
STATEMENT
REASON
B
F triangle with right angle at E
1. nDEF is a right
A
16
D
C
26
16
15°
Ewith right angle at H
2. nGHI is a right triangle
A
D
10
C
16
B HG, choose a point A
3. On the ray opposite
such that HA =16ED.
26
3. Postulate 6-Ruler-Second
Assumption: To every point on a
line, there corresponds exactly
one real number, called the
unique distance between the
points.
16
x°
50°
10 12 D
16
A
2x
C
12
D
A B C D
40°
y°
C
12
50°
40°
B C>D ED
4.A HA
A
12
A
12
D
2. Given
y°
x°
2x
1. Given
B
C
4. __________
40°
50°
12
12
2x
5. Draw AI
50°
A
5. __________
40°
6.
A
/GHI
A
12
B
and /AHI are supplementary angles
C A B
2x
A B C D
A
6. __________
D
E
F
Part F – Congruence - Part 2
417
7. m/GHI + m/AHI = 180
7. __________
8. m/GHI = 90
8. __________
9. 90 + m/AHI = 180
9. __________
10. m/AHI = 90
10. Properties of Algebra
11. /AHI is a right angle
11. Definition of a right angle
12. /AHI > /DEF
12. __________
13. EF > HI
13. __________
14. nAHI > nDEF
14. __________
15. AI > DF
15. __________
16. DF > GI
16. __________
17. AI > GI
17. Transitive Property of Congruent
Line Segments (or substitution)
18. In nIGA, /G > /A
18. Theorem 33: If two sides of a
triangle are congruent, then the
angles opposite them are
congruent.
19. /A > /D
19. CPCTC
20. /G > /D
20. Transitive Property of Congruent
Angles (or substitution)
21. nDEF > nGHI
21. __________
418
Unit IV – Triangles
E
D
H
G
B
5. In the figure to the right, BA = BC, CE = CD,
/EDF is a straight angle, and m/E = 15°.
Find m/ACB, m/A, m/B, and m/BFD.
F
C
D
A
A
I
F
I
F
E
15°
A
6. Using the diagram to the right, find:
a) m/A
b) m/CBD
c) m/ABC
D
2x
B
D
16
2x
F
F y°
x°
A
2x
D
10
H
G
26
16
A B C D
A B C D
H
G
E
E
I
I
C
16
E
D
G
E
D
12
C
G
D
7. nABC is isosceles with AB = BC. If AB =A 4x
find AB and BC.
B Cand
D BC = 6x – 15,
B 40°
50°
B
12
H
H
12
50°
40°
F 12
B DF = 3x + 15,
8. nDEF is isosceles with base DF. If DE = 4x +A 15, EF =FA2x + 45,
and
C
D
A
C
find the lengths of the sides of the triangle.
D 15°
A
E
15°
2x
EA
B
B
26 + 30, find
9. In nXYZ, XY = YZ. If m/X = 4x + 60, m/Y = 2x + 30, and16m/Z16= 14x
16
26
16
m/X, m/Y, and m/Z.
C
A
2x
A B C D2x
10. Using the diagram to the right, find:
a) m/BDC
b) m/DAC
c) m/DBA
d) m/ACB
A
B
D
D
10
16
A B C D
A B C D
Prove:
F
C
2x
50°
12
50°
C
12
D
D 12
D
40°
C
40°
E
12
12
50°
12
A
40°
A
A
40°
50°
12 B
12
A
D
A
2x
nACE > nADB
C
A
A
C
C
A B C D
A B C D
A
A
C
B
B
2x
AC > AD
CB > DE
C
16 y°
A
A
11. Given:
D
y°
E
x°
10
x°
B
E
B
B P
D
D
E
T
F
S
F
DR
M
N
D
2x
E
419
Part F2x – Congruence
E - Part 2
E
A
F
B
C
A B C D
A
A
50
40°
40°
40°
A 50° 1212
A12
B
B 12
50°
40°
12. Given:
AB > AE
AC > AD
A
2x
2x
A
A B
12
A
A
2x
Prove:
nACE > nADB
C
13. Given:
Prove:
A B C
A B C DD
A
A
nAEB is isosceles
DA
AC
AC
FC
>
>
C
B
C
D
D
AB > AC
/A > /CBD
D
DB
A M
A
Y
A
B
P
BD CD
S
B
M
M
Prove:
nMNP with MP > NP
RT
PN
RS
PM
>
>
RT • RM = RS • RN
A
T
P
R
R
S
C
D
A
D
D H
T
4
C
Q
B
B
N
N
T
P
R
S F
EM
H
H
N
P
S
E
EM
A
A
C
C
C
C
B
Prove: AB = BC
15. Given:
C
D
1
2
A
B
X
A
3
14. Given:
E B
10
X
F
D
E
D
PAA
2
F
F
E
2x
D
D
E
E
2x
2x
C
AD • AF = CF • BD
E
B
A B C D
A
B
B
C
F
N
T
B
F R
G
G
N
B
C
G
A altitude
C
16. Prove: A triangle is isosceles if and only if a medianDandAan
of the triangle are
the same segment.
A
B
B
B
420
Unit IV – Triangles
C
C
C
A
A B C A
D
A
A
A
A
12
2x
x°
C
12
x°
x°
12 x°
x°
B
x°
6
C
x°
x°
47°
47°
For Exercises 7-9, use the information
on each figure
to state any valid
16
A B BC indicated
C X
BC
2x
C
D2x
conclusions you can, about the unmarked sides and anglesx° in the figure.
7.
A B C D
A B C D
A
8.
x°
A
6
12
16
C
B
47°
A16
X
X
47° 47°
x°
X
Y
x°20°
H
H
C D
C
A BI C D
6
8
2x
80°
20°
I B 80°
20°
3
K
I
9.
47°
A
6
x°
47°
20°
K
100
3
12 Y
8
J
8
80°
x°
8
K
A
C
100
16
50°
K
8
C
x°
X
10.
11.
12.
13.I
14.
Find BF
H
Find EC 8
80°
20° FE
Find
Find 100
CF
3
8
Find m/B
J
47°A
J
16
16
X
A
x°
16 50° x° 9
16
K
x°
F
x°
50°
E
2x
J
2x
A
9
x°
4
8
50°
50°
C
4
B
50°
R
K
7
M 40°
40°
40°
A
A B C
D 16
A B C
D
15.E Find MR
50°
16. Find9 m/NML
D
17. Find LN
R
M 40°
40°
L
7
PC
B 7
2x
40°
4
C
50°
ML 40° F 40° M50°40°
40°
P
P
E 40°
2
N
D 1
40°A
50°
9
A
N2x
N
R
M 40°
L
A B C D
/A > /C.
D 1
40°
B
A
C
A BD C 1D
C
2x
2x
7
2
E
R
P
2 E= 2x + 745, and
40° a given triangle nABC,
D +1 25,
18. In
If AB2 =E4x
A BC
B
A
M 40°
40°
AC = 3x – 15, find the lengths of all three sides.
L
P
40°
N
A
C
2x
1
D
19. Given:
A
Prove:
A B C D
A B C D
A
A B C DA
A
2
E
/ADE
> /BED
B
DC > EC
A
2x
N
C
2x
2x
D
2x
2x
A
B C D
A
A B C D
A
2x
424
Unit IV – Triangles
B
B
A
2x
E
D
C
2x
C D
7
9
R 50°
x°
L
A
A
4
D
L
D
J
A B
C
P
R
D Use this illustration to complete
NExercises 15-17.
In the figure below,BM is the midpoint of CLP.
A
F
2x
9
9
Y
B
50°
3 9
A
50°
47°
50° 2x 9
50°
D
50°
E
50°
50° C
H
B
B F
8
4
4E
80°
C
C
50°
50°
50°
AF B 20°
50°
C D
9
E I
D
100
50°
9
2x
16
F
6
Y
9
100 x°
100
J
Use nABF
and 47°
nCED in the figure below
to complete
Exercises
10-14.50°
3
3
47°
8
8
16
8
80°
I Y
H
x°
x°
H
6
47°
16
x°
x°
2x
C
C
Y
1
2
E
B
4. /PRQ is a right angle
4. __________
5. nPQR is a right triangle
5. __________
6. Draw nPQR with PR = b and RQ = a
6. Postulate 6 - Ruler - Second
Assumption – To ever pair of
points on a line, there
corresponds exactly one real
number, called the unique
distance between the points.
7. (PQ)2 = a 2 + b 2
7. __________
8. (PQ)2 = c 2
8. __________
9. PQ = c
9. Properties of Algebra
10. AB = PQ; AC = PR; CB = RQ
10. __________
11. AB > PQ; AC > PR; CB > RQ
11. __________
12. nABC > nPQR
12. __________
13. /C > /R
13. __________
14. m/C = m/R
14. __________
15. m/C = 90
15. __________
16. /C is a right angle
16. __________
17. nABC is a right triangle
17. __________
432
Unit IV – Triangles
B
P
B
C
Db
B
R
a
A
D
3
4
Q
C
5
B
C
2
2
4a
B
B
B
3 8
CA
14
8
2.5
C
A
16
13
2
+b =c
2
B
7.5
2
20
A
A
16
320 = 320
B
8217+ 162 > 142
2
82 + 162 < 20
14
8
A
6419.5
+ 256 < 400
B
C
320 < 400
10
D 6
64 + 256 > 196
15
C
A
16
5
A
16
a2 + b2 > c2
a2 +B18b2 < c2
2x
2
2 12
x
8 8 + 16 = 320
3
2.5
C 64 + 256 = 320 A
A B C D
320 > 196
A
B
C
2x
Decrease c
C
A
C
16
4
4
20
CA B C D
A A
b
Increase c
B
C
85
x
B 8 3
A
A
Original
2x
C
B
16
4
A
B
C
5
B
Extending Your
11 x Mind:
20
2x
B
3
8
C
8
Consider the diagrams
below,
showing
the
result
of
increasing
or
decreasing
the length of
x
24
B
C
A
the hypotenuse c, of right nABC.
2
C
A
16
A
D
x
A
25
61
A
16
4
3
C
B
C
5
2x
5
5
B
B
B 18
These
diagrams
suggest
the following: If c is decreased, while keeping
a and b the same,
E
7.5
61
14
17
2
2
2
10
11
a + b 8will be greater than c , and m/C will be
less than 90° (acute).
If c is increased,
D
DC 6
A
15
A 2
B
5
2
2
19.5
a and b xthe same, a + b will be less than c , and m/C will be greater than
while keeping
C
A
C
A
16
C
90° (obtuse). Summarizing these Bideas, we can state:A
B
B
5
D
A
5
B17
A
C
8
5
15
C
8
A
A
16
A
C
B
Use these
ideas in Exercises 11-19 to classify each triangle with the given side lengths as
acute,
right,8or obtuse.
3
B
C
BE
A
C
3
If a 2 + b 2 = c 2 then, the triangleB is a right triangle
E the triangle is an acute triangle
If a 2 + b 2 >A c 2 then,
10
D C6
11
C
If a 2 + b 2 < c 2 Dthen,
the triangle
is an obtuse triangle
11. 4,
D 5,117
C
20
8
A
14.
C
2, 3, 4
C
AC
16
17. 9, 10, 12
C
3, 2 , 5
13. 6, 8, 10
15.
3,
16. 0.4, 0.5, 0.6
4, 5
18. 0.9, 4.0, 4.1
B
B
12.
19.
1 4
, ,1
5 5
14
8
C
A
16
For Exercise 20, decide if nABC is right, acute, or obtuse. Explain your answer.
B
20.
17
A
10
15
D 6
C
A
B
5
E
C
D
Part F – Congruence - Part 2
435
C
G
I
H
95
7
D
6
Exterior Angle of a Polygon – The
angle formedB when one side 47
of a polygon
is
38
A
8
A
E
D of
B
extended. For example, in the diagram below, /BCD is an exterior angle
F
C
nACB.
B
Remote
Interior
Angles
Exterior
Angle
B
C
A
C
D
Note: An exterior angle of a triangle forms a linear pair with
B
A the adjacent
A
X
interior angle of the triangle. Further, the two angles of the triangle that are not
adjacent to the exterior angle are called the remote interior angles of that
B
C
exterior angle.
C
Y
B
A
C
A
Z
Lesson 1 — Exercises:
C
C
3
4
1. Prove Theorem 37 – “If you have a given
exterior angle of a triangle, then the
7
measure of that angle is greater
the measure
of either remote
B
5
A interior angle.”
1 than
8
9
A
2
6
(Exterior Angle Inequality Theorem) (Note: This is the same theorem we proved in
the lesson. We are using it as an exercise to make sure you understand its proof.
C
C
Use your course notes to check.)
C
a) State the theorem.
b) Draw and label a diagram to accurately show
the conditions of the theorem.
D
B
A
A
c) List the given information.
B
d) Write the statement we wish Ato prove.
C
C
B
e) Demonstrate the direct proof using
C the two-column format.
B
B
E
2.
A
Complete the following demonstration of Theorem 37 which does not use
A
B
A
Theorem 26. (Note: It will be helpful for you to draw an appropriate diagram to see the
C
C
C
logic of the proof.)
STATEMENT
1. nABC with exterior /BCD
REASON
B
A
A
B
F
C
A
1. Given
D
E
2. Theorem
4: __________
2. Choose point M on BC so that point M is the
midpoint of BC
F
B
A
3. Draw AE through point M
3. Postulate 2 – Uniqueness of
Lines,
Planes,
and Spaces
J
I
H
C
E
B
A
G
A
C
5
B
437
Part G – Congruence
- Part 3
7
F
C
A
F
B
6
B
B
4. Choose point E on AE so that AM = EM
4. Postulate 6 – Ruler – Second
Assumption
5. Draw CE
5. Postulate 2 – Uniqueness of
Lines, Planes, and
Spaces
6. BM = MC
6. __________
7. BM > MC
7. __________
8. /AMB > /EMC
8. __________
9. AM > EM
9. __________
10. nAMB > nEMC
10. __________
11. /ABM > /ECM
11. __________
12. m/BCD = m/DCE + m/ECM
12. __________
13. m/BCD = m/DCE + m/ABM
13. __________
14. m/BCD > m/B
14. Definition of “is greater than” –
For any numbers a and b,
a > b, if and only if, there is a
positive number c such that
a=b+c
|| EC
15. __________
16. /BAC > /DCE
16. __________
17. m/BAC = m/DCE
17. __________
18. m/BCD = m/DCE + m/ECM
18. __________
19. m/BCD = m/BAC + m/ECM
19. __________
20. m/BCD > m/BAC
20. __________
15. AB
438
Unit IV – Triangles
3. Referring to the figure to the right, use Theorem 37 to determine which angle is larger,
or state that this cannot be determined from the information given.
a. /1 and /6
1
A
b. /1 and /2
2
6
c. /2 and /4
d. /5 and /2
3
4
2x
5
A B C D
C
y°
D
For Exercises 4-7, write an inequality that showsx° a relationship between only x and y.
4. x + 30 = y
b°
c°
A
16
B
5. 75 – yC= 180 – x
x°
6. 60 + 45 + x = 180 + y
7. 46 + x + 89 = 91 + y
1
2
3
A
B
For Exercises 8-11, write an inequality that shows a relationship
between only m/1 and m/2.
r°
16
8. m/1 = 46 + m/2
s°
9. u°25 + 30 + m/1 = m/2
x°
v°
10. 175 + m/1 – 89 = m/2 – 46
t°
x°
11. m/1 + m/3 = m/2
– m/4
1
A
2x
x°
A
12. Given:
b=x+c
3
4
2x
y>x
35°
Cw°
y°
A B C D
A
5
y°
A
x°
c°
A
16
2x
16
m/1 > m/3
C
x°
x°
A
E
m/1 > m/2
b°
C y°
D
D
b°
B
1
D
C
c°
A
2
Prove:
6
3
4
x°
13. Given:
1
5
2
A A
B B
C D
C D
2x
Prove:
2
6
40°
C
2
B
B
3
A D
B
1
1
B
A B C D
2
A
16
B
u°
v°
r°
16
2x
3
C
s°
B
x°
A
3
r°
t°
x°
s°
x°
2x
A
1
u°
D
v°
C
t°
x°
x°
Part G – Congruence
- Part 3
40°
Q
2x
x°
439
The “Hinge” Theorem (Side-to-Angle version) – “If two sides of one triangle
are congruent to two sides of a second triangle, and the length of the third side
of the first triangle is greater than the length of the third side of the second
triangle, then the measure of the angle opposite the third side of the first
triangle, is greater than the measure of the angle opposite the third side of the
second triangle.”
Lesson 2 — Exercises:
1. Theorem 38: “In a given triangle, if two sides are not congruent, then the angles
opposite those sides are not congruent.” (Note: This is the same theorem we proved
in the lesson. We are using it as an exercise to make sure you understand its proof.
Use your course notes to check.)
a) State the theorem.
b) Draw and label a diagram to accurately show the conditions of the theorem.
c) List the given information.
d) Write the statement we wish to prove.
e) Demonstrate the direct proof using the two-column format.
2. Prove: If the measure of one side of a triangle is greater than the measure of a
second side of the triangle, then the measure of the angle opposite the longer side is
greater than the measure of the angle opposite the shorter side. Use the same
outline as used in Exercise 1.
C
5
7
3. In each of the following, the measure of the sides 5of nABC6 are given. List the angles
1
A
of nABC from the smallest to the largest.
9
A
B
7
a) AB = 17
BC = 21
AC = 18
5
1
2x
b) AB = 15
BC = 17
AC = 16
Q
C
8
C
5 6
7
c) AB = 9
BC = 40
AC =A5 41
B C D7
5
5 10 10 6
1
6
d) AB 5= 13 6 BC = 12
AC = 15 A
2
6
A
9
9
A
9
MA
B
7
7
NB
7
8
5
4. Name the largest and the smallest angle
in2xeach of the following figures.
1
16
2x
W Q
Q
a)
b)
C
8
A B
56 C D 7 x° 5
109 10
A B C D
A
5
10
M A
8
6
2x
W
c)
A B C
D x°
9 10
UM
16
x°
1
Q
7
1
5
H
9
10
10 1 2
9
5
Part
13
4
1310
8
2x14
9
6
1 7
8
V
I
10
9
5
2
2
5
10
5
9
2
8
1 2
443
10
8
9
1
G8 – 9Congruence
- Part 3
4
J
2
8
6
9
2
10
5
5
13
A
1
V
5
10 1 2
9
I
8
J
H
x°
8
6
8 9
9
W
9
U2
5
16
N V
813
x°
d)
2
8
6I
10
13
16
6
NV
9
MU
8
W
9
9
U
2
5
5
2
2
1
66
2
9 6
6
9
66
138
H
7
N B
7
16
1
6
10
5
14
10
9
20
13
1
14
75°
MH
C
5
A
5
16
1
x°
6
7
10
5
8
2
5
1 2
10
I
W
2
9
7
N
98
10 9
9
In Exercises 5 and 6, state
the
that can be drawn from the6 given information.9
A
B
7 conclusion(s)
13
5 J
1
Give a reason for each conclusion.
16
V
U
13
1
2x
Q
C
A B
5. Given: C D
nAMR with RA10> AM
5
A
5
7
6
9
16
2x
5
6 5
7
M
6
1
9
7
B
Q
9
6
A
In
N
9
V
I
8
10
1 992
5
1
16 1
13
x° 6
16
4
V
9
10 10 9
5
2x
J
M8
8 14
M
A B C D
20
13
1
9
2
1
20
75° 14
H A5
10
Z
6
7
M
V
C
9
16
A 9
F
8
I
1
2x
2A
75°
20
4
9
10
5
75°
8
x°
9
14
A 5B 9C D2x
920
JY1
13
B C10D
10
10 A
1 2
A
9
14
9 60°
B
8
7
C
A
Z
9
15
777°
1
2
8
85°
15
15
4
7
2
2
8 1 E
13
32°
1X
2
74
10
D
Q
5 7
6
2
4
B
A
C
14
15
5 7
9
D
75°
B
A
R
1
2x
7
1
60°
57°
10
E
8
Z
R
M
S
D
Y
X
8
8
8
A
D
E
B
38°
14or =.
13
For Exercises
12
and
13,
fill
in
the
blank
with
>,
<,
75°
40°
C
1 J
75°
10
77°
15 2 85°
7
M
16 10 1 2
15
2x
9 15
4 15
D
I Y
F
C
9
20
Z
A
A x°
M
2
75°
1
A
A
10
7
12.
13.
A B C D
Q DC < DA
A
4
NOTE:
7
R
2
20 9
F
B
X 5 7
9
Y
14
28°
9
13 1
10
L 32°
N 20
6
A
2x 2
7
60°
57°
Y
X
75°
D 78 E
2x
1
ARAB C 8D 6 B
7
M
E
8
1 D
240°
4
7
Y
P
24°
8
D 3 F
C
B
38°
4
40°
A
C
2
E
A 75°
1
7 A9
M 4
6 89
20
D
Z
2
F
20
S
Z
9
9
9
7
1A
75°
10
A BC
C
5 7
A B C D m/1
D
1 ____28°m/2
85°
6
77°N
M
2L
2x 15
57°
60° 15
15
15
B 7
32° 2x
9
8
8
A
Dm/1____m/2
E
B
2 5 7 775°
6
A
7
N
57°
A B C D
Y
Q
X
B
A2
m/3____m/4
20
1 R
X
P
8
7
E
D 3 7
Y
B
A
4
10
4
M
m/5____m/6
A
4
40°
1 24°
C2
F
6 8
7
AD 1
S
A4Z B
1
2
3
5 7
28°
L 32°
2
N
C
D 9 15 8 85° E
Y
6
77°
6 89
8
15
7
2x
S
38°
15
155 7
C
40°
C 7
B
6
D
7
2x
N
7
57°
C 60°
4
9A B C D
85°
D
F 15
C1
444
Unit
2
9
Q 8 E
P
8
B
A
BIV –DTriangles
R
D 3
B
1AX
4
A
Y
A
2
1
2
6 8
B
B B
Z A
AD 3
S 7
R
4
5 7
4
20
4
6 8
20
6
A
3
N
Y
P
6
7
1
S
4M
8
L 7 28°
10
D
10
9
5
N8
57°
P B
8
10
7
F
9
7 75°
75°
E
Y
9
1
4
21
202
2
8
9.
75°
7
5
75°
60°
2
M
D
28°
9 32°
2
20I
7
57°
9
V
L
1
M 8
F
6
Z
8
1
20
Y
4
20
2
8
X
13
13
6
C
6
A
U
11.
10
19 2
8
6
M
6
2
92x
10
1
8.
H
W
9Z
75°
8 8
x°
9
1
A
X
W
5
4
A
2
5 9
UR
16
10
2
9
10
2 109
angle is larger, /1 or /2.
14
75°
10 20
5
9
5
9
6
7
4
8
Q R
1 2
13
2 N
M1 A 8B C8
D 2x
9
8N
8
9
7
10.
6
2R
10A 5
7
5
1
2
9 5
1
B8
X
10
5
QI
2
A B6 C 10
x°10 1 2 D
2x
V
J
2x
A B C D 10
B
139 6
6
A
27
5
9
Z
5
1
2x
10
U 55
1
7
A
9
16
2
1
7
Exercises
A 2 7-11,7tell which
B
H
8
8
x° 9
A
5
5
7
10
5
7.
6
C
8
V
13
N
75°
13 2
9
A B 6C
D
109
1
J
16
8
20 1 2
M
10
6x°
H
5
6
A
9
2x
m/M
< _____
W
I
A B C D
m/A
8 6 __ _____
2
16
Conclusion:
5
x° 1
6
5
C8
10
I
R5
1
6
6
2x
2
9
7 9
82
9
9
5
14
A
H
10
6
5
5
9
AR < RM
U10
RM5 > MA
8
6
2
W
Q
28
7
N
8
>B _____
x°
10
A B C
6.D 1Given:
8
5
6
1
10 M
7 /M
9
5
A 2
Conclusion:
1
x°
10
2
3
4
6 8
C
B
H
3
9
6
7
10
I
6 8
4
1 2
10
N
14
8
10
For Exercises 14-18, use the
J two triangles shown
16
to the right, and determine whether each
A
statement is always x°true, sometimes
true,
P
or never true.
13
R
5 7
2
1
9
4
3
M
T
S
20
6 8
14.
15.
16.
17.
18.
If
If
If
If
If
NP =
NP =
NP =
NP =
NP >
75°
1
RT, MP =2xTS, Rand NM > RS, then
M m/P < m/T.
E
RT, NM = RS, and PM > ST, then m/N > m/R.2 N
20
Z
RT, NM = RS, then m/N
= m/R.
A
C D = RS, and PM = ST, then m/P = m/S.
RT,B NM
A
P
RT, NM = RS, and PM = ST, then m/P < m/T.
D
75°B
10
C
M
Y
X
R
7
19. Given:
EB >A AE
CD > BC
2x
Prove:
F
C
9
9
60°
57°
m/A > m/DA
D
B
8
E
4
D
1
2
B
6
7
C
A
E
8
T
S
A
M
7
28°
32°
L
5 7
N
20. Prove, or find a counterexample: “The sum of the measures of any two angles of a
7
B C D than the measure of the third angle.”
triangle isAgreater
1
2
P
D
A
Z
2x
15
3
C
85°
15
15
15
21. Write an indirect proof for the following:
Q
X
Y
|| CD
Given:
ABCD is a trapezoid with AB
Prove:
8 both
E right angles
D not
/C and /D are
8
38°
A
R
B
C
40°
9
B
6 8
S
77°
4
D
9
C
A
B
A
24°
B
40°
C
D
7
4
1
2
6
7
Part G – Congruence - Part 3
5 7
445
4. In each of the following, the measures of the angles of nABC are given. List the sides
of nABC from the longest to the shortest.
a.
b.
c.
d.
m/A =
m/A =
m/A =
m/A =
46
9
60
47
m/B
m/B
m/B
m/B
=
=
=
=
30
70
59
48
m/C
m/C
m/C
m/C
=
=
=
=
104
101
61
85A
D
28° 42°
78°
A
A
D
E
29°
78°
28° 42°
83°
A can be drawn from the givenC information.
In Exercises 5 and 6, state the conclusions that
D
B A
2x
28° 42°
E
Give a reason for each conclusion.
83°
78°
29°
A B C DA
5. Given:
/R > /A
Conclusion:
AM > _____
A
B
2x
C
E
83°
29°
A B C D
2x
A B C D
C
B A
R
16
M
A
x°
6. Given:
x°
16
m/A > m/C
m/C < m/B
M
B
R
C
16
M
B
R
C
x°
A
C
Conclusion:
BC > _____
AB __ _____
B
ZA
16
Z A
x°
16
Z
X
x°
right.16
Y
7. Using a ruler, trace the triangle at the
2x
Y
X C
a. Measure XY and XZ.
x°
F
b. Choose a point K on your paper away from the
Y
2x
X
9 C
9
triangle you traced, so that it corresponds to
F
A KMN
B C D where XY60°
point X. Now, draw a new triangle
= KM, XZ = KN,
57°
2x
A 9 8C
B D 9 8 F E
and m/K < m/X.
A B C D
60°
57°
c. Measure ZY and NM. Which is longer?
A 9 8 M B D 9 8
E
d. How do the measures of ZY and ANM
to the m/X
and m/K?
7
B Crelate
60°
57°
A D
28° of
e. Complete this statement of the Angle-to-Side
version
“If two
B Hinge
E
D
8 M The
L A 32°
N8 Theorem:
7 second triangle and the
sides of one triangle are congruent to two sides of a
7
A
M
28°
measure of the included angle of the first2x triangle
greater
than the
L is
N measure of
32°7 P
the included angle of the second triangle,
then ______________.”
A
728°
L 32°
N
2x
Z
f. Why is this theorem referred to as the “Hinge” theorem?
S
P
15
2x
A B C D
448
Unit IV – Triangles
X
15
A
A B C D
X D15
BA
XD
7
77°
Z
77°
Z
15
15
P
15
Y
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S
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40°
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D
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15 E
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15
77°60° 2x
158
15 A
15
7
S28° A P
N28°
N
32°
A B C D
7S
Y
B
28°
P
2x
N the blank with >, <. or
85°
In Exercises
8 - 14, fill in
P32°
X
Y Q
15
15
2x
A B C D
C7
Z
M
Z
F
R
X
Y Q
S
2x
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7
A
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S B AC D
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8.
9.
10.
X
Y
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D
8
77°
E
77°
A
15
85°
A B28°C D
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15 85° 15
15
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9 85°
77°9
15
L 38°
N
40°
A
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8
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15
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A B C D
8
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8
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9
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BCM____ EF
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8
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S
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8
8 7 E
38°
A AB C D
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40°
8
C
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9
77°
28°
9
40°
85°
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L 32°
15A
15
9
9
15
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9
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A
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8
7
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40°
A
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B
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R
X
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A
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8
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C
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40°
D 40°
40°
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BC
____
AD
BD
____
CD
C
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40°
15
15
9
24°
A 15
15
24°
12
12
7
X
L
A
7
24°
A B C D
X
40°
B
Q
D
2x
A
12
C
A
47°
C
X
40°
24°
X
14
Z
Z
Y
58°
12C
12
Y
14
C
X
58°
12
14
14
Y
14
16. Given:
C 47°
A
Z
Y
47°
14
14
Z
Y
X Y
D
>
>
AD Y CB
AB
DAC
C
B
A
B
D
A
T
BC >D AD
B
X
D
Y
T
A
2 1
B
Y
D
C
A
Y
14
Prove:
14
Z 47°
C
12
Z
X
X
47°
X
X
Prove: 47° YZ > XZ
ZY
14
Z 12
X
Y
58°
YZ 12> XY
X
C 14
12 {In other words, prove that the
Y
Z
C
12
C
hypotenuse
of a right triangle47°is D
the
C
14
longest side of the triangle.}
X
C
C
12
Y
X
12
Y
12
58°
Y
14
X
12
12
X
C
14
14
12
B47° D /YXZ is
C a right angle
15. Given:
14
Y
47°
12
12
12
12
12
D
C
D
12
X24°
B
CB
12
58°
12
14
14
D
12
40°
B 12
C58°
12
12
B B
C
A
12
14. X __?__
YC
X
C
D 8
38°
40°
12
12 9
58° 9
12
12 12
58°
12
13.D C __?__
8B
E 12
A
A
R
C
24°
R
58° 9
2x
Y
15
C
B
A
T
B
2 1
A
B
T
X
1
2 T
C
Y
R
AR
D
U
3
4
3
4
2 1
T
U
T
S
1
1
U
Part G –2 Congruence
- 2Part
3
S
R
A
3
4B
S
449
Example 1: The lengths of two sides of a triangle are 6 and 8. Find the possible length, x,
for the third side.
Solution:
x > |8 - 6|
x >2
and
x < |8 + 6|
x < 14
or 2 < x and x < 14
2 < x < 14
So the length of the third side is between 2 and 14.
Lesson 4 — Exercises:
1. Theorem 40: “In a given triangle, the sum of the lengths of any two sides, is greater
than the length of the third side.” (Note: This is the same theorem we proved in the
lesson. We are using it as an exercise to make sure you understand its proof. Use
your course notes to check.)
a) State the theorem.
b) Draw and label a diagram to accurately show the conditions of the theorem.
c) List the given information.
d) Write the statement we wish to prove.
e) Demonstrate the direct proof using the two-column format.
2. Can the sum of the lengths of two side of a triangle equal the length of the third side?
Explain why or why not.
3. In nABC, AB + BC > AC. Draw triangle ABC and write two other inequalities.
In Exercises 4-9, tell whether it is possible for a triangle to have sides of the given lengths.
Assume that all given lengths are positive. Justify your answers.
4. 9, 7, 14
5. 4, 3, 7
6. 3, 4, 5
7. 18, 10, 7
8. x, 2, x – 2
(where x is a
natural number)
9. 2x, x, x + 1
(where x is a
natural number)
452
Unit IV – Triangles
Unit IV — Triangles
Appendix A – Properties of Real Numbers
Properties of the Real Numbers — This term refers to the postulates, or axioms, which
we accepted without proof, in the study of Arithmetic. Following is a comprehensive list for
your reference.
1. Properties of Relations:
• Trichotomy - For any real numbers a and b, only one of the following can be true:
a = b, a > b, a < b.
• Reflexivity for Equality - For any real number a, a = a.
• Symmetry for Equality - For any real numbers a and b, if a = b, then b = a.
• Transitivity for Equality - For any real numbers a, b, and c, if a = b, and b = c,
then a = c.
• Substitution - For any real numbers a, and b, if a = b, then a can be substituted
for b in any expression (or b can be substituted for a).
• Transitivity for Inequality - For any real numbers a, b, and c, if a > b, and b > c,
then a > c. Likewise, if a < b, and b < c, then a < c.
2. Properties of Well-Defined Operations:
• Existence - For any real numbers a and b, a + b, a - b, and a · b exist. Further,
a 4 b exists, as long as b = 0.
• Uniqueness - For any real numbers a and b, a + b, a - b, and a · b are unique.
Further, a 4 b is unique, as long as b = 0.
• Closure - For any real numbers a and b, a + b, a - b, and a · b are real numbers.
Further, a 4 b is a real number, as long as b = 0.
3. Properties of Operations in General:
• Commutativity of Addition - For any real numbers a and b, a + b = b + a.
• Commutativity of Multiplication - For any real numbers a and b, a · b = b · a.
• Associativity of Addition - For any real numbers a, b, and c,
(a + b) + c = a + (b + c).
• Associativity of Multiplication - For any real numbers a, b, and c,
(a · b) · c = a · (b · c).
• Distributivity of Multiplication over Addition - For any real numbers a, b, and c,
a · (b + c) = a · b + a · c.
• Distributivity of Multiplication over Subtraction - For any real numbers a, b,
and c, a · (b - c) = a · b - a · c.
A-1
Unit IV – Triangles
Arc of a Circle (p. 158) – Any set of continuous points on a circle.
Area (p. 43, p. 48, p. 52, p. 55, p. 59, p. 62) – Intuitively, the number of non-overlapping
unit squares and parts of unit squares, which will exactly cover the interior of a simple
closed plane curve
Area of a Circle (p. 62, p. 76) – The product of the square of the radius of the circle, and
the irrational number p.
Area of a Parallelogram (p. 48) – The product of the base and height of a parallelogram.
Area of a Rectangle (p. 43) – The product of the base and height of a rectangle.
Area of a Regular Polygon (p. 59) – One-half the product of the measure of the apothem,
and the perimeter of the polygon.
Area of a Rhombus (p. 48) – The product of the base and height of a rhombus.
Area of a Square (p. 43) – The square of the measure of one side of a square.
Area of a Triangle (p. 52) – One-half of the product of the base and height of a triangle.
Area of a Trapezoid (p. 55) – One-half of the product of the height and the sum of the
bases of a trapezoid.
Auxiliary Element (p. 291, p. 351, p. 365, p. 404, p. 426, p. 430, p. 442, p. 446, p. 451) –
From a Latin word meaning “to help”, this is a geometric element which is not specifically
referred to in a given diagram, but which does exist, and is needed, in order to logically
complete the demonstration of a conditional. This element is usually drawn in a figure,
using dashed lines, and must be referenced in the proof with a step justifying its existence.
It is important to note that you must not place too many conditions on the element (called
“over-determining”) thereby denying its existence. Neither must you place too few
conditions on the element (called “under-determining”), thereby allowing the existence of
more than one such element. In other words, a geometric element is considered to be
“determined”, if exactly one such element can be drawn to meet the conditions.
Base Angles of an Isosceles Triangle (p. 320) – The angles opposite the legs in an
isosceles triangle.
Base of an Isosceles Triangle (p. 320) – The third side of an isosceles triangle, when the
triangle has exactly two congruent sides.
Betweenness of Points (p. 124) – For three collinear points, A, B, and C, with
coordinates a, b, and c, respectively, point B is said to lie between points A and C, if and
only if, either a < b < c, or, a > b > c.
B-2
Unit IV – Triangles
Disjunction (p. 92) – An operation in logic which joins two simple statements, using the
word “or”.
“Divides Proportionally” (p. 351) – A term referring to points on line segments which
“divide” the segments into smaller segments which are in the same ratio.
Diagonal of a Polygon (p. 377) A line segment joining any two non-consecutive vertices
of the polygon.
Diagonal of a Rectangular Solid (p. 377) – A line segment joining any two vertices in a
rectangular solid which are not vertices of the same face. Vertices of this type are called
“opposite” vertices.
Diameter of a Circle (p. 62, p. 158) – A line segment which is a chord of a circle, and
passes through the center of that circle.
Dihedral Angle (p. 142, p. 307) – The union of two half-planes (called faces), with the
same edge.
Dilation (p. 8) – A transformation in which the distance between any point of the preimage, and a specified point (called the center of dilation), is multiplied by some constant
factor, to produce the image.
Direct Proof (p. 215) – The process of reaching a desired conclusion, logically and
deductively, from given statements, from already accepted definitions and postulates, and
from any previously proved theorems.
Discrete Geometry (p. 25) – A Geometry in which every point is a “dot”, and every line is
made up of separate points, with a space between them.
Disjoint Sets (p. 18) – Two or more sets which have no members, or elements, in
common.
Dodecagon (p. 34) – A polygon made with twelve line segments.
Edge (p. 141) – Another name for a separation line in a plane.
Element of a Set (p. 18) – Also referred to as a “member” of a set, this is one of the
objects in a set.
Empty Set (p. 18) – A set which contains no members.
Equal Angles (p. 151) – Two angles whose measures are equal.
Equal Line Segments (p. 134) – Line segments whose lengths are equal.
B-6
Unit IV – Triangles
Interior Angles on the Same Side of a Transversal (p. 277) – Pairs of angles which are
between parallel lines and are on the same side of a transversal.
Interior of an Angle (p. 141) – The set of points between two rays when one ray lies in
the edge of a half-plane.
Intersecting Lines (p. 128) – Two lines which have a point in common.
Intersection (p. 18) – An operation on two or more sets, which selects only those
elements common to (or belonging to) all of the original sets.
Intuition (p. 79, p. 81) – A type of mental activity which gives information or beliefs, based
on hunches or insight.
Inverse (p. 107) – A conditional which results from negating both the hypothesis and the
conclusion in a given conditional.
“Is Greater Than” (p. 442, p. 446, p .451) – By definition, for any real numbers a and b, a
“is greater than” b, if and only if, there is a positive real number c, such that, a = b + c.
“Is Less Than” (p. 442, p. 446, p. 451) – By definition, for any real numbers a and b, a “is
less than” b, if and only if, there is a positive real number c, such that, b = a + c.
Isosceles Trapezoid (p. 33) – A trapezoid in which the two non-parallel sides are
congruent.
Isosceles Triangle (p. 33, p. 320, p. 415, p. 422) – A triangle is an isosceles triangle, if
and only if, it has at least two congruent sides.
Isometry Transformation (p. 8, p. 382) – A transformation or combination of
transformations which results in the image being exactly the same shape and size as the
pre-image.
Kite (p. 33) – A quadrilateral in which there are two distinct pairs of consecutive sides
which are of equal measure.
Law of Syllogism (p. 101) – An application of syllogistic reasoning involving two related
conditionals, which, if considered together, using the “law of detachment”, will result in a
third valid conditional.
Legs of an Isosceles Triangle (p.320) – The two congruent sides of an isosceles triangle,
when that triangle has exactly two congruent sides.
Length of a Line Segment (p.134) – A real number which represents the distance
between the endpoints of a line segment.
Appendix B - Definitions and Important Terms
B-9
Line (p. 3) – A basic element of Geometry, which has infinite length, but no thickness. In a
drawing, it is represented by a line segment with arrowheads on each end, to show that it
goes on forever. We name a line by choosing any two points on the line, and labeling
each with a capital letter.
Line Segment (p. 134) – The union of two points on a line, and the set of all the points
between them
Linear Pair (p. 151, p. 316, p. 327) – Two angles which have a common side (they are
adjacent), and whose exterior sides are opposite rays.
Logic (p. 92, p. 100, p.107) – A system of reasoning, in an orderly fashion, which draws
conclusions from specific premises.
Major Arc of a Circle (p. 159) – An arc which is the union of two points on a circle, not the
endpoints of a diameter, and the set of points on the circle which lie in the exterior of the
angle formed by the radii containing the two points.
Mapping (p. 7) – Another name for a transformation in Geometry.
Means-Extremes Product Property of a Proportion (p. 336) – A property of a valid
standard proportion, which states that the product of the means is equal to the product of
the extremes.
Means of a Proportion (p. 331) – In a standard proportion, the second and third terms.
Measure of a Dihedral Angle (p. 307) – A real number which is defined to be the
measure of any of its plane angles.
Measure of a Major Arc of a Circle (p. 159) – A real number which is equal to 360 minus
the measure of its related minor arc.
Measure of a Minor Arc of a Circle (p.159) – A real number which is equal to the
measure of its related central angle.
Measure of the Arc making up a Complete Circle (p. 159) – Related to a central angle
of 360° (a complete rotation about the center of a circle), this defined to be 360°.
Median in a Triangle (p.316, p.416, p.423) – A segment drawn from a vertex of a triangle,
to the midpoint of its opposite side.
Midpoint of a Line Segment (p. 135, p. 239) – A point on a line segment which is
between the endpoints, and divides the given segment into two congruent segments.
B-10
Unit IV – Triangles
Unit IV — Triangles
Appendix C - Postulates and Postulate Corollaries
Postulate 1 – Existence of Points (WT- p. 170)
- “Every line contains at least 2 different points.”
- “Every plane contains at least 3 different, non-collinear points.”
- “Space contains at least 4 different, non-coplanar points, no three of which are
collinear.”
Postulate 2 – Uniqueness of Lines, Planes, and Spaces (WT- p. 175)
- “For any 2 different points, there is exactly 1 line containing them.”
- “For any 3 different, non-collinear points, there is exactly 1 plane containing them.”
- “For any 4 different, non-coplanar points, no 3 of which are collinear, there is
exactly 1 space containing them.”
Postulate 3 – One, Two, and Three Dimensions (WT- p. 178)
- “For any 2 different points in a plane, the line containing them is in the plane.”
- “For any line in a plane, there is at least 1 point in the plane that is not on the line.”
- “For any plane in space, there is at least 1 point in space that is not on the plane.”
Postulate 4 – Separation of Lines, Planes, and Spaces (WT- p. 180)
- “A point separates a line into two non-empty sets called half-lines. If two points
are in the same half-line, then the segment joining them does not contain the
given point. If two points are in different half-lines, then the segment joining them
does contain the given point.”
- “A line separates a plane into two non-empty sets called half-planes. If two points
are in the same half-plane, then the segment joining them does not intersect the
given line. If two points are in different half-planes, then the segment joining them
does intersect the given line.”
- “A plane separates space into two non-empty sets called half-spaces. If two points
are in the same half-space, then the segment joining them does not intersect the
given plane. If two points are in different half-spaces, then the segment joining
them does intersect the given plane.”
Postulate 5 – Intersection of Lines or Planes (WT- p. 184)
- “If 2 different lines intersect, the intersection is a unique point.”
- “If 2 different planes intersect, the intersection is a unique line.”
Appendix C - Postulates and Postulate Corollaries
C-1
Postulate 6 – Ruler (WT- p. 187)
- “The set of all points on a line can be put into a one-to-one correspondence with
the real numbers, so that any point may correspond to 0, and any other point may
correspond to 1.”
- “To every pair of points on a line, there corresponds exactly one real number,
called the unique distance between the points.”
- “The distance between any 2 points on a line is the absolute value of the
difference between their coordinates.”
- “If, on a line, point B lies between points A and C, then:
mAB + mBC = mAC
(Segment-Addition Assumption)
Postulate 7 – Protractor (WT- p. 194)
- “In a half-plane, the set of all rays with a common endpoint in the edge of the
half-plane, can be put into a one-to-one correspondence with the real numbers
from 0 to 180, inclusive, pairing either ray in the edge of the half-plane with 0.”
- “To every pair of rays with a common endpoint in the edge of a half-plane, there
corresponds exactly one real number from 0 to 180, inclusive, called the unique
measure of the angle formed by the rays.”
- “The measure of an angle is the absolute value of the difference between the
coordinates of its rays.”
- “If, in a half-plane, a ray OB lies between rays OA and OC, then:
(Angle-Addition Assumption)
m/AOB + m/BOC = m/AOC
Postulate 8 – Circle (WT- p. 198)
- “The set of all points on a circle can be put into a one-to-one correspondence with
the real numbers from 0 to 360, inclusive, with the exception of any one point
which may be paired with 0 and 360.”
- “To every pair of points on a circle, there correspond exactly 2 real numbers whose
sum is 360, each of which may be called the distance between the 2 points.”
- “The distance between any 2 points on a circle is the absolute value of the
difference between their coordinates.”
- “If, on a circle, a point B lies between points A and C, then:
mAB + mBC = mAC
(Arc-Addition Assumption)
Postulate 9 – Uniqueness of Parallel Lines (WT- p. 202)
- “In a plane, through a point not on a given line, there is exactly one line parallel to
the given line.”
Postulate 10 – Uniqueness of Perpendicular Lines (WT- p. 207)
- “In a plane, through a point not on a given line, there is exactly one line
perpendicular to the given line.”
- “Through a point not in a given plane, there is exactly one line perpendicular to the
given plane.”
Postulate 11 – Corresponding Angles of Parallel Lines (WT- p. 277)
- “If two parallel lines are cut by a transversal, then corresponding angles are
congruent.”
C-2
Unit IV – Triangles
Postulate Corollary 13b (WT- p. 395)
- “If the hypotenuse and an acute angle of one right triangle are congruent to the
hypotenuse and the corresponding acute angle of another right triangle, then the
two right triangles are congruent.” (HA Congruence Postulate Corollary)
Postulate Corollary 13c (WT- p. 395)
- “If a leg and an acute angle and of one right triangle are congruent to the
corresponding leg and acute angle of another right triangle, then the two right
triangles are congruent.”
(LA Congruence Postulate Corollary)
Postulate Corollary 13d (WT- p. 395)
- “If the two legs of a right triangle are congruent to the two corresponding legs of
another right triangle, then the two right triangles are congruent.”
(LL Congruence Postulate Corollary)
Postulate Corollary 13e (WT- p. 395)
- “If the hypotenuse and one leg of one right triangle are congruent to the
hypotenuse and the corresponding leg of another right triangle, then the two right
triangles are congruent.”
(HL Congruence Postulate Corollary)
C-4
Unit IV – Triangles
Theorem 18 (WT- p. 287)
“If a given line is perpendicular to one of two parallel lines, then it is perpendicular
to the other.”
Theorem 19 (WT- p. 291)
“If two lines are cut by a transversal so that corresponding angles are congruent,
then the two lines are parallel.”
Theorem 20 (WT- p. 294)
“If two lines are cut by a transversal so that alternate interior angles are congruent,
then the two lines are parallel.”
Corollary 20a (WT- p. 294)
“If two lines are cut by a transversal so that alternate exterior angles are congruent,
then the two lines are parallel.”
Theorem 21 (WT- p. 297)
“If two lines are cut by a transversal so that interior angles on the same side of the
transversal are supplementary, then the two lines are parallel.”
Corollary 21a (WT- p. 297)
“If two lines are cut by a transversal so that exterior angles on the same side of the
transversal are supplementary, then the two lines are parallel.”
Theorem 22 (WT- p. 301)
“If two lines are perpendicular to a third line, then the two lines are parallel.”
Theorem 23 (WT- p. 304)
“If two lines are parallel to a third line, then the two lines are parallel to each other.”
Theorem 24 (WT- p. 307)
“If two parallel planes are cut by a third plane, then the two lines of intersection are
parallel.”
Theorem 25 (WT- p. 324)
“If you have any given triangle, then the sum of the measures of its angles is 180.”
Corollary 25a (WT- p. 324)
“If two angles of one triangle are congruent to two angles of another triangle, then
the third pair of angles are congruent.”
Corollary 25b (WT- p. 324)
“If all of the angles of a triangle are congruent, then the measure of each angle
is 60.”
Appendix D - Theorems and Theorem Corollaries
D-3
Corollary 30a (WT- p. 364)
“If you have a right triangle, then either leg is the geometric mean between the
hypotenuse of the triangle, and the projection of that leg on that hypotenuse.”
Corollary 30b (WT- p. 364)
“If you have a right triangle, then the altitude drawn to the hypotenuse of that
triangle is the geometric mean between the segments of that hypotenuse, formed
by drawing that altitude.”
Corollary 30c (WT- p. 364)
“If you have an altitude drawn to the hypotenuse of a right triangle, then the product
of the lengths of that altitude and the hypotenuse, is equal to the product of the
lengths of the two legs.”
Theorem 31 (WT- p. 369)
“If you have a right triangle, then the square of the measure of the hypotenuse, is
equal to the sum of the squares of the measures of the two legs.”
(The Pythagorean Theorem)
Corollary 31a (WT- p. 369)
“If you have a right triangle whose acute angles have measures of 30° and 60°,
then the measure of the hypotenuse is twice the measure of the shorter leg, and the
measure of the longer leg is 3 times the measure of the shorter leg.”
Corollary 31b (WT- p. 369)
“If you have a right triangle whose acute angles each have measures of 45°, then
the measure of the hypotenuse is 2 times the measure of either leg.”
Theorem 32 (WT- p. 414)
“If two given triangles are both congruent to a third triangle, then the two given
triangles are congruent to each other.”
Theorem 33 (WT- p. 415)
“If two sides of a triangle are congruent, then the angles opposite them are
congruent.”
Corollary 33a (WT- p. 416)
“If a triangle is equilateral, then it is equiangular.”
Corollary 33b (WT- p. 416)
“If a triangle is equilateral, then the measure of each of its angles is 60O.”
Theorem 34 (WT- p. 422)
“If two angles of a triangle are congruent, then the sides opposite them are
congruent.”
Appendix D - Theorems and Theorem Corollaries
D-5