Elementary Geometry for College Students
6th Edition
ISBN:9781285195698
Author:Daniel C. Alexander, Geralyn M. Koeberlein
Publisher:Daniel C. Alexander, Geralyn M. Koeberlein
Chapter10: Analytic Geometry
Section10.CT: Test
Problem 13CT: Using as few variables as possible, state the coordinates of each point if DEF is isosceles with ...
Related questions
Question
![4-6 Triangle Congruence: ASA, AAS, and HL
EXAMPLES
EXERCISES
1 Given: B is the midpoint of AE.
ZA = ZE,
16. Given: Cis the midpoint
of AG.
HA || GB
H
ZABC = ZEBD
Prove: AABC=AEBD
Prove: AHAC= ABGC
B
17. Given: WX 1 XZ,
YZ 1 ZX,
WZ = YX
Proof:
Statements
Reasons
Prove: AWZX=AYXZ
Z,
1. Given
2. Given
3. Given
4. Def. of mdpt.
5. ASA Steps 1, 4, 2
1. ZA = ZE
2. ZABC E ZEBD
W
3. B is the mdpt. of AE.
4. AB = EB
5. ΔΑΒC= ΔΕBD
18. Given: ZS and ZV
R
are right angles.
RT = UW.
mZT= mZW
Prove: ARST :AUVW
4-7 Triangle Congruence: CPCTC
EXAMPLES
EXERCISES
1 Given: JL and HK bisect each other.
19. Given: M is the midpoint
of BD.
BC = DC
Prove: ZJHG= ZLKG
K
Prove: 21 = 2
B
M
D
20. Given: PQ= RQ,
PS = RS
Prove: QS bisects ZPQR.
Proof:
Statements
Reasons
1. JI and HK bisect
1. Given
each other.
2. JG = LG, and
HG = KG.
2. Def. of bisect
R
3. ZJGH = LLGK
3. Vert. A Thm.
21. Given: H is the midpoint of GJ.
L is the midpoint of MK.
GM = KJ, GJ = KM,
G
4. AJHG = ALKG
5. ZJHG = ZLKG
4. SAS Steps 2, 3
5. СРСТС
ZG= ZK
Prove: ZGMH= ZKJL
M L
K
298
Chapter 4 Triangle Congruence](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3785ad64-93eb-4008-8832-b47a1a40c0c9%2Fe9727ef9-1e58-4a09-af42-ffb262c61d25%2Fwvtko6d_processed.jpeg&w=3840&q=75)
Transcribed Image Text:4-6 Triangle Congruence: ASA, AAS, and HL
EXAMPLES
EXERCISES
1 Given: B is the midpoint of AE.
ZA = ZE,
16. Given: Cis the midpoint
of AG.
HA || GB
H
ZABC = ZEBD
Prove: AABC=AEBD
Prove: AHAC= ABGC
B
17. Given: WX 1 XZ,
YZ 1 ZX,
WZ = YX
Proof:
Statements
Reasons
Prove: AWZX=AYXZ
Z,
1. Given
2. Given
3. Given
4. Def. of mdpt.
5. ASA Steps 1, 4, 2
1. ZA = ZE
2. ZABC E ZEBD
W
3. B is the mdpt. of AE.
4. AB = EB
5. ΔΑΒC= ΔΕBD
18. Given: ZS and ZV
R
are right angles.
RT = UW.
mZT= mZW
Prove: ARST :AUVW
4-7 Triangle Congruence: CPCTC
EXAMPLES
EXERCISES
1 Given: JL and HK bisect each other.
19. Given: M is the midpoint
of BD.
BC = DC
Prove: ZJHG= ZLKG
K
Prove: 21 = 2
B
M
D
20. Given: PQ= RQ,
PS = RS
Prove: QS bisects ZPQR.
Proof:
Statements
Reasons
1. JI and HK bisect
1. Given
each other.
2. JG = LG, and
HG = KG.
2. Def. of bisect
R
3. ZJGH = LLGK
3. Vert. A Thm.
21. Given: H is the midpoint of GJ.
L is the midpoint of MK.
GM = KJ, GJ = KM,
G
4. AJHG = ALKG
5. ZJHG = ZLKG
4. SAS Steps 2, 3
5. СРСТС
ZG= ZK
Prove: ZGMH= ZKJL
M L
K
298
Chapter 4 Triangle Congruence
![4-8 Introduction to Coordinate Proof
EXAMPLES
EXERCISES
1 Given: ZB is a right angle in isosceles right
AABC. E is the midpoint of AB.
D is the midpoint of CB. AB = CB
Position each figure in the coordinate plane and give
the coordinates of each vertex.
22. a right triangle with leg lengths r and s
Prove: CE = AD
23. a rectangle with length 2p and width p
Proof: Use the coordinates A(0, 2a) , B(0,0),
and C(2a, 0). Draw AD and CE.
24. a square with side length 8m
For exercises 25 and 26 assign coordinates to each
vertex and write a coordinate proof.
25. Given: In rectangle ABCD, E is the midpoint of
E
AB, Fis the midpoint of BC, Gis the
midpoint of CD, and His the midpoint
of AD.
B
D
Prove: EF= GH
By the Midpoint Formula,
(0 +0 2a+0
26. Given: APQR has a right ZQ.
Mis the midpoint of PR.
E =
= (0, a) and
2
(0 + 2a 0+ 0
D=
|= (a, 0)
Prove: MP = MQ= MR
2
27. Show that a triangle with vertices at (3, 5), (3, 2),
and (2, 5) is a right triangle.
By the Distance Formula,
CE = V(2a – 0)² + (0 – a)²
=V4a? + a² = a/5
AD=V(a – 0)² + (0 – 2a)²
a² + 4a² = a/5
Thus CE = AD by the definition of congruence.
4-9 Isosceles and Equilateral Triangles
EXAMPLE
EXERCISES
1 Find the value of x.
Find each value.
E
mZD+ mZE+ mZF= 180°
28. х
M
K
(45 – 3x)°
by the Triangle Sum
Theorem. mZE = mZF
by the Isosceles
Triangle Theorem.
D
42°
mZD+2mZE= 180°
Substitution
29. RS
R
42 + 2(3x) = 180
Substitute the given
values.
2у — 4.5
1.5y
6x = 138
Simplify.
x = 23
Divide both sides by 6.
30. Given: AACD is isosceles with ZD as the vertex
angle. Bis the midpoint of AC.
АВ — х+ 5, ВС3 2х — 3, and CD 3D 2x + 6.
Find the perimeter
AACD.
Study Guide: Review
299](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3785ad64-93eb-4008-8832-b47a1a40c0c9%2Fe9727ef9-1e58-4a09-af42-ffb262c61d25%2Fit70kir_processed.jpeg&w=3840&q=75)
Transcribed Image Text:4-8 Introduction to Coordinate Proof
EXAMPLES
EXERCISES
1 Given: ZB is a right angle in isosceles right
AABC. E is the midpoint of AB.
D is the midpoint of CB. AB = CB
Position each figure in the coordinate plane and give
the coordinates of each vertex.
22. a right triangle with leg lengths r and s
Prove: CE = AD
23. a rectangle with length 2p and width p
Proof: Use the coordinates A(0, 2a) , B(0,0),
and C(2a, 0). Draw AD and CE.
24. a square with side length 8m
For exercises 25 and 26 assign coordinates to each
vertex and write a coordinate proof.
25. Given: In rectangle ABCD, E is the midpoint of
E
AB, Fis the midpoint of BC, Gis the
midpoint of CD, and His the midpoint
of AD.
B
D
Prove: EF= GH
By the Midpoint Formula,
(0 +0 2a+0
26. Given: APQR has a right ZQ.
Mis the midpoint of PR.
E =
= (0, a) and
2
(0 + 2a 0+ 0
D=
|= (a, 0)
Prove: MP = MQ= MR
2
27. Show that a triangle with vertices at (3, 5), (3, 2),
and (2, 5) is a right triangle.
By the Distance Formula,
CE = V(2a – 0)² + (0 – a)²
=V4a? + a² = a/5
AD=V(a – 0)² + (0 – 2a)²
a² + 4a² = a/5
Thus CE = AD by the definition of congruence.
4-9 Isosceles and Equilateral Triangles
EXAMPLE
EXERCISES
1 Find the value of x.
Find each value.
E
mZD+ mZE+ mZF= 180°
28. х
M
K
(45 – 3x)°
by the Triangle Sum
Theorem. mZE = mZF
by the Isosceles
Triangle Theorem.
D
42°
mZD+2mZE= 180°
Substitution
29. RS
R
42 + 2(3x) = 180
Substitute the given
values.
2у — 4.5
1.5y
6x = 138
Simplify.
x = 23
Divide both sides by 6.
30. Given: AACD is isosceles with ZD as the vertex
angle. Bis the midpoint of AC.
АВ — х+ 5, ВС3 2х — 3, and CD 3D 2x + 6.
Find the perimeter
AACD.
Study Guide: Review
299
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