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ven: bisects ZAED. B 2EA. the midpoint of CE pve: AEBA - AECD D. Statemente

ven: bisects ZAED. B 2EA. the midpoint of CE pve: AEBA - AECD D. Statemente

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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Ok, so I waited an hour, just to be told it violates the honor code, even though I stressed this is not GRADED. So I going to ask again but stress even more "NOT GRADED" 

 

Instructions: Use all the grey Statements, not all the yellow reasons will be used. Please, understand that the yellow reasons will have a few extra left unused. Provide which statement goes with which reason in order from 1st to last. Please refer to the image attached! 

 

Thanks in advance!

111 X A Class X A Cour X A 151: X O Simil x A Engli X * (3,09 x O Edita X G "Sim x O Simil x Copy x b Mat x G Dear X A https://docs.google.com/presentation/d/1tvB_sBSpGATbST6Rc1kz3EljivEmVMeRamlbptmXmSc/edit#slide=id.gb1c59f0845_4_219 aProct DngDrp on wleroe Capy of Capyof a Via imi .-- spamor n Rroof TRIANGLE 8 NOT ALL REASONS WILL BE USED! Given: SIMILAR CE bisects ZAED. ED = 2EA. Bis the midpoint of CE SSS Similarity B. Substitution Property PROOFS D. Prove: AEBA - AECD E Defn. of Midpoint Directions: Complete the proof to the right using the statements and reasons. Statements Reasons 1. place statement here 1. place reason here Division Property EB + BC = EC 2. place statement here AA Similarity ED = 2EA 2. place reason here 3. place statement here Given 3. place reason here 2EB = EC 4. place statement here 4. place reason here Segment Addiftion Postulate ED2 EA 5. place statement here 5. place reason here Given EC 2- EB 6. place statement here 6. place reason here Simplify ER + ER = EC 7. place statement here 7. place reason here Defn. of Angle Bisector EB = BC 8. place statement here 8. place reason here SAS Similarity B is the midpoint of CE 9. place statement here 9. place reason here CE bisects ZAED 10. place statement here 10. place reason here Division Property 11. place statement here 11. place reason here ZAEB = ZDEC Substitution Property AEBA - AECD 12. place statement here 12. place reason here Given ED EC EA ER B OGina Wison (AlI Things Nigebrat, LC), 201B answers.jpg Results.png Show all a 4:53 PM A O ) J A U 4/16/2021
Transcribed Image Text:111 X A Class X A Cour X A 151: X O Simil x A Engli X * (3,09 x O Edita X G "Sim x O Simil x Copy x b Mat x G Dear X A https://docs.google.com/presentation/d/1tvB_sBSpGATbST6Rc1kz3EljivEmVMeRamlbptmXmSc/edit#slide=id.gb1c59f0845_4_219 aProct DngDrp on wleroe Capy of Capyof a Via imi .-- spamor n Rroof TRIANGLE 8 NOT ALL REASONS WILL BE USED! Given: SIMILAR CE bisects ZAED. ED = 2EA. Bis the midpoint of CE SSS Similarity B. Substitution Property PROOFS D. Prove: AEBA - AECD E Defn. of Midpoint Directions: Complete the proof to the right using the statements and reasons. Statements Reasons 1. place statement here 1. place reason here Division Property EB + BC = EC 2. place statement here AA Similarity ED = 2EA 2. place reason here 3. place statement here Given 3. place reason here 2EB = EC 4. place statement here 4. place reason here Segment Addiftion Postulate ED2 EA 5. place statement here 5. place reason here Given EC 2- EB 6. place statement here 6. place reason here Simplify ER + ER = EC 7. place statement here 7. place reason here Defn. of Angle Bisector EB = BC 8. place statement here 8. place reason here SAS Similarity B is the midpoint of CE 9. place statement here 9. place reason here CE bisects ZAED 10. place statement here 10. place reason here Division Property 11. place statement here 11. place reason here ZAEB = ZDEC Substitution Property AEBA - AECD 12. place statement here 12. place reason here Given ED EC EA ER B OGina Wison (AlI Things Nigebrat, LC), 201B answers.jpg Results.png Show all a 4:53 PM A O ) J A U 4/16/2021
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