Question

given, ab ≅ de, bc ≅ ef, prove: ac ≅ df
(there are line segments: ac with points a, b, c and df with points d, e, f)
statements + justification bank:
definition of congruence, bc = de, ab ≅ de, ab + bc = ac, ac = df
segment addition postulate, definition of congruence, given, given, ab + de = df
ac ≅ df, segment addition postulate, ab = de, bc + ef = df, substitution property of equality
bc ≅ ef, transitive property, ab + bc = de + ef, definition of congruence, substitution property

Explanation:

Step1: Identify Given Information

We know \( AB = DE \) (given) and \( BC = EF \) (given, from the diagram and problem context).

Step2: Apply Segment Addition Postulate

For segment \( AC \), \( AC = AB + BC \) (Segment Addition Postulate: a whole segment is the sum of its parts).
For segment \( DF \), \( DF = DE + EF \) (Segment Addition Postulate).

Step3: Substitute Equal Segments

Since \( AB = DE \) and \( BC = EF \), substitute \( AB \) with \( DE \) and \( BC \) with \( EF \) in the equation for \( AC \):
\( AC = DE + EF \).
But \( DF = DE + EF \) (from Step 2), so by the Transitive Property, \( AC = DF \).

Answer:

\( AC = DF \) (proven using Segment Addition Postulate and Transitive Property with given \( AB = DE \) and \( BC = EF \))