Below are
\[\triangle ABC\] and
\[\triangle DEF\]. We assume that
\[AB=DE\],
\[m\angle A=m\angle D\], and
\[m\angle B=m\angle E\].
Triangle A B C and triangle D E F. Side A B and side D E are congruent. Angle A and angle D are congruent. Angle B and angle E are congruent.
\[A\]
\[B\]
\[C\]
\[D\]
\[E\]
\[F\]
Triangle A B C and triangle D E F. Side A B and side D E are congruent. Angle A and angle D are congruent. Angle B and angle E are congruent.
Here is a rough outline of a proof that
\[\triangle ABC\cong\triangle DEF\]:
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.]
As a result of these transformations,
\[B'\] must coincide with
\[E\]. [Show drawing.]
As a result of these transformations,
\[C'\] must coincide with
\[F\]. [Show drawing.]
What is the justification that
\[C'=F\] in step 3?
Choose 1 answer:
Choose 1 answer:
(Choice A)
\[C'\] and
\[F\] are the same distance from
\[D\] along the same ray.
A
\[C'\] and
\[F\] are the same distance from
\[D\] along the same ray.
(Choice 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

\[\overleftrightarrow{DE}\].
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

\[\overleftrightarrow{DE}\].
(Choice C)
\[C'\] and
\[F\] are at the intersection of the same pair of rays.
C
\[C'\] and
\[F\] are at the intersection of the same pair of rays.