Question

1. b is the midpoint of ac; e is the midpoint of df, and ac ≅ df. given
2. ab = bc and de = ef definition of midpoint
3. ac = df definition of ≅ segments, 1st use
4. ac = ab + bc and df = de + ef segment addition postulate
5. ac = ab + ab and df = de + de substitution property, 1st use
6. ac = 2ab and df = 2de division property of equality
7. ac = 2de substitution property, 2nd use

Explanation:

Step1: Identify the given information

The first statement "B is the midpoint of \(AC\), \(E\) is the midpoint of \(DF\), and \(AC\cong DF\)" is the given information.

Step2: Use the definition of midpoint

Since \(B\) is the midpoint of \(AC\), by the definition of midpoint \(AB = BC\). Similarly, since \(E\) is the midpoint of \(DF\), \(DE=EF\).

Step3: Use the definition of congruent segments

If \(AC\cong DF\), then by the definition of congruent segments (which states that if two segments are congruent, their lengths are equal), \(AC = DF\).

Step4: Apply the Segment Addition Postulate

The Segment Addition Postulate states that for a line - segment \(AC\) with a point \(B\) between \(A\) and \(C\), \(AC=AB + BC\), and for a line - segment \(DF\) with a point \(E\) between \(D\) and \(F\), \(DF = DE+EF\).

Step5: Substitute using the mid - point relationships

Since \(AB = BC\) (from step 2), we can substitute \(BC\) with \(AB\) in \(AC=AB + BC\) to get \(AC=AB + AB\). Similarly, since \(DE = EF\) (from step 2), we can substitute \(EF\) with \(DE\) in \(DF=DE + EF\) to get \(DF=DE + DE\).

Step6: Simplify using the properties of equality

From \(AC=AB + AB\), we can use the distributive property (\(a+a = 2a\)) to get \(AC = 2AB\). Similarly, from \(DF=DE + DE\), we get \(DF = 2DE\).

Step7: Substitute using the equality from step 3

Since \(AC = DF\) (from step 3), and \(AC = 2AB\), \(DF = 2DE\), we can substitute \(DF\) with \(AC\) (or vice - versa). If \(AC = DF\) and \(DF = 2DE\), then \(AC=2DE\) (by substitution).

Answer:

1. \(B\) is the midpoint of \(AC\), \(E\) is the midpoint of \(DF\), and \(AC\cong DF\)
2. \(AB = BC\) and \(DE = EF\)
3. \(AC = DF\)
4. \(AC=AB + BC\) and \(DF = DE+EF\)
5. \(AC=AB + AB\) and \(DF=DE + DE\)
6. \(AC = 2AB\) and \(DF = 2DE\)
7. \(AC=2DE\)