Given: FD bisects EFC Steps 1. FD bisects«EFC; FC Reasons FC bisects «DFB Given bisects «DFB D 2. KEFD = «DFC , DFC = «CFB of Definition of Congruent Transitive Prop 3. B 4. m
Given: FD bisects EFC Steps 1. FD bisects«EFC; FC Reasons FC bisects «DFB Given bisects «DFB D 2. KEFD = «DFC , DFC = «CFB of Definition of Congruent Transitive Prop 3. B 4. m
Holt Mcdougal Larson Pre-algebra: Student Edition 2012
1st Edition
ISBN:9780547587776
Author:HOLT MCDOUGAL
Publisher:HOLT MCDOUGAL
Chapter6: Ratio, Proportion, And Probability
Section: Chapter Questions
Problem 6CR
Related questions
Question
![9. Given: FD bisects «EFC
Steps
1. FD bisects «EFC; FC
Reasons
FC bisects <DFB
Given
bisects DFB
E
2. KEFD = «DFC,
DFC = «CFB
Definition of Congruent
Transitive Prop
3.
B
4. m<EFD= m«CFB
5. KEFD = <CFB
Prove: *EFD=«CFB](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F934fbfaf-6c69-4f0b-836e-8605b68f5540%2F9e6d3a69-990e-466f-ab15-f10b6406f1b0%2Fbp6aym_processed.jpeg&w=3840&q=75)
Transcribed Image Text:9. Given: FD bisects «EFC
Steps
1. FD bisects «EFC; FC
Reasons
FC bisects <DFB
Given
bisects DFB
E
2. KEFD = «DFC,
DFC = «CFB
Definition of Congruent
Transitive Prop
3.
B
4. m<EFD= m«CFB
5. KEFD = <CFB
Prove: *EFD=«CFB
![10. Given: <WXY is a right angle
Steps
1. <WXY is a right angle
Reasons
Given
Y
2. mgWXY =90°
2
3.
3. mx2+m<3 = M<WXY
4. mx2+mx3=90°
Substitution
5. 女1=A3
Given
Prove: mx2+mx1=90°
6. m¤1=m<3
7. mx2+m1=90°
Substitution](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F934fbfaf-6c69-4f0b-836e-8605b68f5540%2F9e6d3a69-990e-466f-ab15-f10b6406f1b0%2Fr30zxqx_processed.jpeg&w=3840&q=75)
Transcribed Image Text:10. Given: <WXY is a right angle
Steps
1. <WXY is a right angle
Reasons
Given
Y
2. mgWXY =90°
2
3.
3. mx2+m<3 = M<WXY
4. mx2+mx3=90°
Substitution
5. 女1=A3
Given
Prove: mx2+mx1=90°
6. m¤1=m<3
7. mx2+m1=90°
Substitution
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