Question

here is a rough outline of a proof that $\triangle abc \cong \triangle def$: 1. we can map $\triangle abc$ 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 $\triangle abc$ across $\overleftrightarrow{de}$ and then $c = f$, $a = d$ and $b = e$. show drawing. what is the justification that $c = f$ in step 3? choose 1 answer: a $c$ and $f$ are the same distance from $e$ along the same ray. b both $c$ and $f$ lie on intersection points of circles centered at $d$ and $e$ with radii $df$ and $ef$, respectively. there are two such possible points, one on each side of $\overleftrightarrow{de}$. c $c$ and $f$ are at the intersection of the same pair of rays.

Explanation:

To justify \( C'' = F \) in step 3, we analyze the options:

• Option A: Focuses on distance from \( E \), but the key is about the intersection of circles (or rays) from \( D \) and \( E \), so A is incorrect.
• Option B: Describes circles centered at \( D \) and \( E \) with radii \( DF \) and \( EF \), but after reflection, we use the fact that \( C'' \) and \( F \) lie on the same pair of rays (from \( D \) and \( E \)) rather than just circle intersections with two points. So B is incorrect.
• Option C: After reflecting \( \triangle A'B'C' \) across \( \overleftrightarrow{DE} \), \( A'' = D \) and \( B'' = E \). So \( C'' \) lies on the ray from \( D \) through \( A''(=D) \) (which is consistent with \( F \)’s position relative to \( D \) and \( E \)) and the ray from \( E \) through \( B''(=E) \). Thus, \( C'' \) and \( F \) are at the intersection of the same pair of rays (from \( D \) and \( E \)), making this the correct justification.

Answer:

C. \( C'' \) and \( F \) are at the intersection of the same pair of rays.