Dear beloved readers, welcome to our website! We hope your visit here brings you valuable insights and meaningful inspiration. Thank you for taking the time to stop by and explore the content we've prepared for you.
------------------------------------------------ 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.
