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below are △abc and △def. we assume that ab = de, bc = ef, and ac = df. …

Question

below are △abc and △def. we assume that ab = de, bc = ef, and ac = df. there is a rough outline of a proof that △abc ≅ △def: 1. we can map △abc using a sequence of rigid transformations so that a = d and b = e. (show drawing.) 2. if c and f are on the same side of (overleftrightarrow{de}), then c = f. (show drawing.) 3. if c and f are on opposite sides of (overleftrightarrow{de}), then we reflect △abc across (overleftrightarrow{de}). then c = f, a = d and b = e. (show drawing.) what fact can we use to justify step 1? choose 1 answer: a (ab = de) and segments with the same length are congruent. b (bc = ef) and segments with the same length are congruent. c (ac = df) and segments with the same length are congruent.

Explanation:

Brief Explanations

Step 1 is about mapping \( \triangle ABC \) so that \( A = D \) and \( B = E \). The key here is the segment \( AB \) and \( DE \). We know \( AB = DE \) (given), and segments with the same length are congruent. So the fact that justifies step 1 is that \( AB = DE \) and congruent segments (same length) can be mapped via rigid transformations to coincide. Option A states \( AB = DE \) and segments with the same length are congruent, which matches the reasoning for step 1. Options B and C refer to \( BC = EF \) and \( AC = DF \) respectively, which are not the segments involved in mapping \( A \) to \( D \) and \( B \) to \( E \).

Answer:

A. \( AB = DE \) and segments with the same length are congruent.