College

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.

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

Answer :

Mapping △ABC to align with DE:

Translate △ABC so that point A coincides with point D.

Rotate △ABC about point A (which now coincides with D) until side AB aligns with side DE.

Label the resulting triangle as A'B'C'.

Drawing:

Sketch △ABC and △DEF as given.

Show the translation of △ABC to A'B'C', with A' overlapping D and AB aligned with DE.

C' and F on the same side of DE:

If C' lies on the same side of line DE as point F, then segments C'F and AC' (which is equal to DF) form a congruent triangle with segments DF and AC.

By the Side-Side-Side (SSS) Congruence Postulate, △ABC is congruent to △DEF.

Drawing:

Extend lines AC' and DF to show their intersection.

Mark the angles and sides that prove congruence.

C' and F on opposite sides of DE:

If C' lies on the opposite side of line DE from point F, reflect triangle A'B'C' across DE.

Label the reflection as A"B"C".

Since reflection is a rigid transformation, A"B"C" is congruent to A'B'C', and therefore also to △ABC.

Now, C" will coincide with F, and A" and B" will still coincide with D and E, respectively.

Again, by SSS Congruence, △ABC is congruent to △DEF.

Drawing:

Show the reflection of A'B'C' across DE to form A"B"C".

Mark the coinciding points and corresponding sides to demonstrate congruence.

Conclusion:

In both cases (C' and F on the same or opposite sides of DE), we've established that △ABC is congruent to △DEF through a sequence of rigid transformations, proving that ∠ABC ≅ ∠DEF.