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.

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.