Question

below are $\triangle abc$ and $\triangle def$. we assume that $ab = de$, $m\angle a = m\angle d$, and $m\angle b = m\angle e$.
two triangles, $\triangle abc$ (teal) and $\triangle def$ (red), with right angles at $b$ and $e$ respectively
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 $a = d$, $b$ and $e$ are on the same ray from $d$, and $c$ and $f$ are on the same ray from $d$. show drawing.
2. as a result of these transformations, $b$ must coincide with $e$. show drawing.
3. as a result of these transformations, $c$ must coincide with $f$. show drawing.

what is the justification that $b = e$ in step 2?
choose 1 answer:
a) $b$ and $e$ are the same distance from $d$ along the same ray.
b) both $b$ and $e$ lie on intersection points of circles centered at $d$ and $f$ with radii $de$ and $fe$, respectively. there are two such possible points, one on each side of $\overleftrightarrow{df}$.
c) $b$ and $e$ are at the intersection of the same pair of rays.

Explanation:

To determine the justification for \( B' = E \) in step 2, we analyze each option:

• Option A: Since we mapped \( \triangle ABC \) such that \( A' = D \), \( B' \) and \( E \) are on the same ray from \( D \), and \( AB = DE \) (given), \( B' \) and \( E \) must be the same distance from \( D \) along that ray. By the uniqueness of a point at a given distance from a vertex on a ray, \( B' \) coincides with \( E \).
• Option B: This describes a construction for triangle congruence (SSS or SAS - like circle intersections) but is not relevant here as we are using rigid transformations and ray - based mapping, not circle intersections.
• Option C: \( B' \) and \( E \) are on the same ray from \( D \), not the intersection of two rays (the other ray is for \( C' \) and \( F \)), so this is incorrect.

Answer:

A. \( B' \) and \( E \) are the same distance from \( D \) along the same ray.