Below are 
\[\triangle ABC\] and 
\[\triangle DEF\]. We assume that 
\[AB=DE\], 
\[BC=EF\], and 
\[AC=DF\].
Triangle A B C and Triangle D E F. Side A B and side E D have congruent signs. Side B C and side E F also have congruent signs. Side A C and side D F also have congruent signs.
\[A\]
\[B\]
\[C\]
\[D\]
\[E\]
\[F\]
Triangle A B C and Triangle D E F. Side A B and side E D have congruent signs. Side B C and side E F also have congruent signs. Side A C and side D F also have congruent signs.
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 
\[A'=D\] and 
\[B'=E\]. [Show drawing.]
If 
\[C'\] and 
\[F\] are on the same side of 
 
\[\overleftrightarrow{DE}\], then 
\[C'=F\]. [Show drawing.]
If 
\[C'\] and 
\[F\] are on opposite sides of 
 
\[\overleftrightarrow{DE}\], then we reflect 
\[\triangle A'B'C'\] across 
 
\[\overleftrightarrow{DE}\]. Then 
\[C''=F\], 
\[A''=D\] and 
\[B''=E\]. [Hide drawing.]
Triangle D E F with point A prime mapped onto point D. B prime is mapped onto point E. C double prime is mapped onto point F. A dashed line extends through points D and E in both directions. The triangle is reflected across the line where A double prime is mapped onto A prime and D, B double prime is mapped onto B prime and E. C prime is equal distance away from point F on the other side of the dashed line.
\[C'\]
\[A''=A'=D\]
\[B''=B'=E\]
\[C''=F\]
Triangle D E F with point A prime mapped onto point D. B prime is mapped onto point E. C double prime is mapped onto point F. A dashed line extends through points D and E in both directions. The triangle is reflected across the line where A double prime is mapped onto A prime and D, B double prime is mapped onto B prime and E. C prime is equal distance away from point F on the other side of the dashed line.
What is the justification that 
\[C''=F\] in step 3?
Choose 1 answer:
Choose 1 answer:
(Choice A, Checked) 
\[C''\] and 
\[F\] are the same distance from 
\[E\] along the same ray.
A
\[C''\] and 
\[F\] are the same distance from 
\[E\] 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.
                                             
                                          
                                          
                                       