Question

1. e is the midpoint of \\( \overline { d f } \\)
2. \\( d e = e f \\)
3. \\( d e + d e = d e + e f \\)
4. \\( 2 d e = d e + e f \\)
5. \\( d e + e f = d f \\)
6. \\( 2 d e = d f \\)

2 given: \\( \overline { k l } \cong \overline { l n } , \overline { l m } \cong \overline { l n } \\)
prove: \\( l \\) is the midpoint of \\( \overline { k m } \\)

1. \\( \overline { k l } \cong \overline { l n } , \overline { l m } \cong \overline { l n } \\)
2. \\( k l = l n , l m = l n \\)
3. \\( k l = l m \\)
4. \\( l \\) is the midpoint of \\( \overline { k m } \\)

Explanation:

Step1: Analyze the first table

• Statement 3:
• Reason: Addition Property of Equality (if \(a = b\), then \(a + c=b + c\). Here \(a = DE\), \(b = EF\), \(c = DE\)).
• Statement 4:
• Reason: Simplification (\(DE+DE = 2DE\)).
• Statement 5:
• Reason: Segment Addition Postulate (if \(D\), \(E\), \(F\) are collinear points, then \(DE + EF=DF\)).
• Statement 6:
• Reason: Substitution (substitute \(DE + EF\) with \(DF\) in \(2DE=DE + EF\)).

Step2: Analyze the second table

• Statement 1:
• Reason: Given.
• Statement 2:
• Reason: Definition of congruent segments (if \(\overline{AB}\cong\overline{CD}\), then \(AB = CD\)).
• Statement 3:
• Reason: Transitive Property of Equality (if \(a = b\) and \(b = c\), then \(a = c\). Here \(a = KL\), \(b = LN\), \(c = LM\)).
• Statement 4:
• Reason: Definition of mid - point (if a point \(L\) divides a segment \(\overline{KM}\) such that \(KL=LM\), then \(L\) is the mid - point of \(\overline{KM}\)).

Answer:

For the first table:

1. Addition Property of Equality; 4. Simplification; 5. Segment Addition Postulate; 6. Substitution.

For the second table:

1. Given; 2. Definition of congruent segments; 3. Transitive Property of Equality; 4. Definition of mid - point.