Question

write a two - column proof.
if ( g ) is the midpoint of ( overline{ab},overline{cd} ) and ( overline{ef},overline{ag}congoverline{ge} ) and ( overline{gd}congoverline{gb} ), then ( overline{ab}congoverline{cd} ), and ( overline{ab}congoverline{ef} ).

Explanation:

Step1: Definition of midpoint

Since \(G\) is the midpoint of \(\overline{AB}\), \(\overline{AG}\cong\overline{GB}\). Since \(G\) is the midpoint of \(\overline{CD}\), \(\overline{CG}\cong\overline{GD}\). Since \(G\) is the midpoint of \(\overline{EF}\), \(\overline{EG}\cong\overline{GF}\).

Step2: Given information

We are given that \(\overline{AG}\cong\overline{GE}\) and \(\overline{GD}\cong\overline{GB}\).

Step3: Transitive property

From \(\overline{AG}\cong\overline{GE}\) and \(\overline{EG}\cong\overline{GF}\), we get \(\overline{AG}\cong\overline{GF}\) (by transitive property of congruence).
From \(\overline{GD}\cong\overline{GB}\) and \(\overline{AG}\cong\overline{GB}\), we get \(\overline{AG}\cong\overline{GD}\) (by transitive property of congruence).

Step4: Length relationships

Since \(\overline{AB}=\overline{AG}+\overline{GB}\), \(\overline{CD}=\overline{CG}+\overline{GD}\), and \(\overline{EF}=\overline{EG}+\overline{GF}\), and \(\overline{AG}\cong\overline{GB}\), \(\overline{CG}\cong\overline{GD}\), \(\overline{EG}\cong\overline{GF}\), \(\overline{AG}\cong\overline{GD}\), \(\overline{AG}\cong\overline{GE}\)
We have \(AB = 2AG\), \(CD=2GD\), \(EF = 2GE\)
Since \(\overline{AG}\cong\overline{GD}\), then \(AB = CD\) (so \(\overline{AB}\cong\overline{CD}\)). Since \(\overline{AG}\cong\overline{GE}\), then \(AB=EF\) (so \(\overline{AB}\cong\overline{EF}\))

Answer:

Statements	Reasons
\(\overline{AG}\cong\overline{GB}\), \(\overline{CG}\cong\overline{GD}\), \(\overline{EG}\cong\overline{GF}\)	Definition of mid - point
\(\overline{AG}\cong\overline{GE}\), \(\overline{GD}\cong\overline{GB}\)	Given
\(\overline{AB}=\overline{AG}+\overline{GB}\), \(\overline{CD}=\overline{CG}+\overline{GD}\), \(\overline{EF}=\overline{EG}+\overline{GF}\)	Segment addition postulate
\(AB = 2AG\), \(CD = 2GD\), \(EF=2GE\)	Substitution (\(\overline{AG}\cong\overline{GB}\), \(\overline{CG}\cong\overline{GD}\), \(\overline{EG}\cong\overline{GF}\))
\(\overline{AB}\cong\overline{CD}\), \(\overline{AB}\cong\overline{EF}\)	Substitution (\(\overline{AG}\cong\overline{GD}\), \(\overline{AG}\cong\overline{GE}\))