Congruent triangles SSS and SAS proofs. Write a 2 column for each of the following problem. No explaining needed just the 2 column proof please.  

Images:
This image has been Flagged as inappropriate Click to unflag
Image (1 of 5)
This image has been Flagged as inappropriate Click to unflag
Image (2 of 5)
This image has been Flagged as inappropriate Click to unflag
Image (3 of 5)
This image has been Flagged as inappropriate Click to unflag
Image (4 of 5)
This image has been Flagged as inappropriate Click to unflag
Image (5 of 5)
Expert Answers
durbanville eNotes educator| Certified Educator

1)R.T.P.  `Delta BAD cong Delta DCB` ` `

                 AB       //    DC                    Given

                 AB       =     CD                    Given (line segments congruent)

       `therefore <ABD = <CDB`              Alternate angles AB // DC  

              BD         =     BD                    Common line

      `therefore Delta BAD cong Delta DCB `             SAS

2) R.T.P. `Delta ABD cong Delta BDC`

             BD          =      BD                   Common line

             AB          =      BC                   Given equilateral `Delta`

             AD         =       DC                  Given equilateral `Delta`

     `therefore Delta ABD cong Delta BDC`              SSS

3) R.T.P. `Delta GDC cong Delta GAF`

           AG         =       GD                   Given G is midpoint of AD

           FG          =      GC                   Given G is midpoint of FC

          `angle FGA = angle DGC`            vertically opposite angles

      `therefore Delta GDC cong Delta GAF`             SAS

4) R.T.P. `Delta BDC cong Delta BDA`

            AB        =       BC                    Given (line segments congruent) 

            AD        =       DC                    Given (BD bisects AC)

            BD        =       BD                    Common side   

       `therefore Delta BDC cong Delta BDA`            SSS

5) R.T.P. `Delta ECD cong Delta BFA`

          `angle DEC = angle ABF`             Given `Delta FGE cong Delta CHB`

             GE       =        BH                  Given `Delta FGE cong Delta CHB`  

             DG       =      HA                    Given (line segments congruent)

           `therefore` DE      =      BA                    Shown above

             FE        =     BC                    Given  `Delta FGE cong Delta CHB`

             FC        =     FC                    Common side

       ` therefore`    EC       =      BF                     Shown above 

       `therefore Delta ECD cong Delta BFA`             SAS 

Ans: 

1) SAS     2) SSS     3) SAS     4) SSS      5) SAS