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7. Given: ml = 45° and mx2 = 45° D. Steps 1. m1 = 45° and m<2= 45° 2. ma1 =m<2 в Reasons Given Substitution %3D 3. 1 2 A Prove: AB is bisector of DAC 4. Definition of angle bisector

7. Given: ml = 45° and mx2 = 45° D. Steps 1. m1 = 45° and m<2= 45° 2. ma1 =m<2 в Reasons Given Substitution %3D 3. 1 2 A Prove: AB is bisector of DAC 4. Definition of angle bisector

Elementary Geometry For College Students, 7e
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
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Prove: m1+mx2+mx3=90° 7. Given: m1 = 45° and m2= 45° Steps Reasons 1. ma1 = 45° and m<2= 45° 2. ma1 =m<2 Given Substitution 3. l= 42 A C 4. Definition of angle Prove: AB is bisector of DAC bisector 8. Given: *HKJ is a straight angle Steps Reasons KI bisects «HKJ 1. «HKJ is a straight angle Given 2. Definition straight angle Given 3. Kİ bisects <«HKJ 4. «IKJ = <IKH 5. mxIKJ = m<IKH K Defintion of congruent Angle Addition Postulate 6. Prove: <IKJ is a right angle 7. m<IKJ +m<IKJ = 180° Substitution 8. 2(m<IKJ) =180° Simplify 9. Division Prop 10. <IKJ is a right angle 9. Given: FD bisects EFC Steps Reasons FC bisects <DFB 1. FD bisects «EFC; FC Given bisects DFB D. 2. «EFD = <DFC , «DFC = «CFB Definition of Congruent Transitive Prop 3. 4. m<EFD=m«CFB 5. «EFD= <CFB B Prove: EFD=<CFB 10. Given: <WXY is a right angle Steps 1. <WXY is a right angle Reasons Given Y 2. M<WXY =90° 3. mx2+m<3 = m<WXY 4. mx2+m<3=90° 5. 1= 43 6. mal = mx3 2 Substitution Given Prove: mx2+mg1=90° 7. mx2+m<l =90° Substitution
Transcribed Image Text:Prove: m1+mx2+mx3=90° 7. Given: m1 = 45° and m2= 45° Steps Reasons 1. ma1 = 45° and m<2= 45° 2. ma1 =m<2 Given Substitution 3. l= 42 A C 4. Definition of angle Prove: AB is bisector of DAC bisector 8. Given: *HKJ is a straight angle Steps Reasons KI bisects «HKJ 1. «HKJ is a straight angle Given 2. Definition straight angle Given 3. Kİ bisects <«HKJ 4. «IKJ = <IKH 5. mxIKJ = m<IKH K Defintion of congruent Angle Addition Postulate 6. Prove: <IKJ is a right angle 7. m<IKJ +m<IKJ = 180° Substitution 8. 2(m<IKJ) =180° Simplify 9. Division Prop 10. <IKJ is a right angle 9. Given: FD bisects EFC Steps Reasons FC bisects <DFB 1. FD bisects «EFC; FC Given bisects DFB D. 2. «EFD = <DFC , «DFC = «CFB Definition of Congruent Transitive Prop 3. 4. m<EFD=m«CFB 5. «EFD= <CFB B Prove: EFD=<CFB 10. Given: <WXY is a right angle Steps 1. <WXY is a right angle Reasons Given Y 2. M<WXY =90° 3. mx2+m<3 = m<WXY 4. mx2+m<3=90° 5. 1= 43 6. mal = mx3 2 Substitution Given Prove: mx2+mg1=90° 7. mx2+m<l =90° Substitution
4. Given: m<A+m<B=90°; <A= C Steps 1. m<A+mB=90° Re 7/18 Given 2. KA E «C 3. m<A= m«C Given 4. Substitution Prove: m<C+m<B=90° 5. Given: <1 and <2 form a straight angle Steps Reasons 1. <1 and <2 form a straight angle Given 2. m<ABC =180° Definition of Straight 1 Angle Angle Addition Postulate 3. 4. Substitution Prove: m<1+ mx2=180° 6. Given: MXEAC=90° Steps 1. m<EAC =90° Reasons Given D 2. Angle Addition Postulate 3. Substitution A
Transcribed Image Text:4. Given: m<A+m<B=90°; <A= C Steps 1. m<A+mB=90° Re 7/18 Given 2. KA E «C 3. m<A= m«C Given 4. Substitution Prove: m<C+m<B=90° 5. Given: <1 and <2 form a straight angle Steps Reasons 1. <1 and <2 form a straight angle Given 2. m<ABC =180° Definition of Straight 1 Angle Angle Addition Postulate 3. 4. Substitution Prove: m<1+ mx2=180° 6. Given: MXEAC=90° Steps 1. m<EAC =90° Reasons Given D 2. Angle Addition Postulate 3. Substitution A
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