Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
Related questions
Question

Transcribed Image Text:M
Given: PM bisects ZLPN
P
1
Prove: 2 41
Statements
Reasons
a. PM bisects LPN
b. 22 23
a.
b.
c. Vertical Angles Are a
с.
d. 22 21
d.
GEOMETRY
Name
Notes Section 2.6 Day 2
Proofs With Definitions
Today we will be looking at proofs which require definitions.
Remember, definitions are always biconditional – they are true both ways!
Definition of a Right Angle
IF and angle's measure is 90°
THEN it is a right angle
OR
IF an angle is a right angle
THEN its measure is 90°
Definition of Complementary Angles
Definition of Supplementary Angles
IF two angles sum to 90
THEN they are complementary
IF two angles sum to 180°
THEN they are supplementary
OR
OR
IF two angles are supplementary
IF two angles are complementary
THEN they sum to 90°
THEN they sum to 180°
1. Given: ZABE is a right angle.
Prove: 24 is complementary to 42
2. Given: 2 is supplementary to 23
Prove: 45 is supplementary to 43
D
A
B
2
3
4
E
Statements
Reasons
Statements
Reasons
1. ZABE is a right
angle
2. MLABE = 90°
1.
1.
1.
2.
2.
2.
3. Angle Addition
Postulate
3.
3.
3.
4.
4.
4.
4.
5. Vertical Angles
Theorem
5.
5.
5.
6.
6.
6.
6.
7. Substitution Property
of Equality
8.
7.
8. 4 is
complementary to 2

Transcribed Image Text:2.6 Day 1 Homework: I can prove statements about segments and angles.
Given: ZLMO S ENMP
Prove: LMPS ZNMO
Statements
Reasons
a.
b. MZLMO = MZNMP
b.
c. MZLMO=mzLMP+m<PMO
C.
MZNMP=mZ NMO +mzPMO
d. mzLMP+MLPMO=m2 NMO +M<PMO
e. mzLMP=mzNMO
f. ZLMP S ZNMO
f.
Given: AC BD
Prove: AB CD
Statements
Reasons
a. AC BD
b. AC = BD
a.
b.
c. AB + BC = AC
C.
d. BC + =
d. Segment Addition Postulate
e. AB + BC = BC + CD
e
f. Subtraction Property
g. Definition of Congruent Segments
f.
g.
Given: B is the midpoint of AC
B
Prove: AB = AC
Statements
Reasons
a. Bis the midpoint of AC
a.
b. AB = BC
b.
с. АВ + ВС AC
с.
d. AB + AB = AC
d.
e. 2AB = AC
e.
f. AB = AC
f.
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