'D Given: ZACB = ZDCB, and BC bisects ZABD. Prove AABC =ADBC Statements Reasons 1 Given ACB = ZDCB 2 BC bisects ZABD 3 ZABC = ZDBC BC=CB 1 4 5 AABC =ADBC 2345

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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What goes in the blank for 2?     [ Select ]   ["Perpendicular", "SAS", "Given", "Angle Addition"] 

         

What goes in the blank for 3?    [ Select ]     ["Reflexive Property", "Alternate Interior Angles", "Segment Bisector", "Definition of Bisector"] 

   

      

What goes in the blank for 4?   [ Select ]    ["CPCTC", "Reflexive Property", "Vertical Angles", "Definition of Bisector"]     

 

    

What goes in the blank for 5?   [ Select ]  ["SAS", "SSS", "AAS", "ASA"]           

'D
C
Given: ZACB E ZDCB, and BC bisects ZABD. Prove AABC =ADBC
Statements
|1 |ACB = ¿DCB
BC bisects ZABD
|3 ZABC = ZDBC
BC=CB
Reasons
1 Given
3
4
4
5
AABC =ADBC
5
Transcribed Image Text:'D C Given: ZACB E ZDCB, and BC bisects ZABD. Prove AABC =ADBC Statements |1 |ACB = ¿DCB BC bisects ZABD |3 ZABC = ZDBC BC=CB Reasons 1 Given 3 4 4 5 AABC =ADBC 5
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