Question

given: • ab ≅ de • ac ≅ df • bc ≅ ef • ∠a ≅ ∠d • ∠b ≅ ∠e • ∠c ≅ ∠f proof: 1. segments ab and de are the same length so they are congruent. therefore, there is a rigid motion that takes ab to de. 2. apply that rigid motion to triangle abc. the image of a will coincide with d, and the image of b will coincide with e. 3. we cannot be sure that the image of c coincides with f yet. if necessary, reflect the image of triangle abc across de to be sure the image of c, which we will call c, is on the same side of de as f. (this reflection does not change the image of a or b.) 4. we know the image of angle a is congruent to angle d because rigid motions dont change the size of angles. 5. c must be on ray df since both c and f are on the same side of de, and make the same angle with it at d. 6. segment dc is the image of ac and rigid motions preserve distance, so they must have the same length. 7. we also know ac has the same length as df. so dc and df must be the same length. 8. since c and f are the same distance along the same ray from d, they have to be in the same place. 9. we have shown that a rigid motion takes a to d, b to e, and c to f; therefore, triangle abc is congruent to triangle def.

Explanation:

Step1: Identify congruent segments

Given $\overline{AB}\cong\overline{DE}$, $\overline{AC}\cong\overline{DF}$, $\overline{BC}\cong\overline{EF}$. Rigid - motion exists for congruent segments.

Step2: Apply rigid motion to $\triangle ABC$

Applying the rigid motion for $\overline{AB}$ to $\triangle ABC$, image of $A$ coincides with $D$ and image of $B$ coincides with $E$.

Step3: Adjust position of $C$'s image

If needed, reflect the image of $\triangle ABC$ across $DE$ so that the image of $C$ (denoted as $C'$) is on the same side of $DE$ as $F$.

Step4: Consider angle - congruence

Since rigid motions preserve angle - size, the image of $\angle A$ is congruent to $\angle D$.

Step5: Locate $C'$ on ray $DF$

$C'$ and $F$ are on the same side of $DE$ and make the same angle with $DE$ at $D$, so $C'$ is on ray $DF$.

Step6: Use distance - preservation of rigid motion

$DC'$ (image of $AC$) and $AC$ have the same length due to rigid - motion distance preservation.

Step7: Equate lengths

Given $AC = DF$, so $DC'=DF$.

Step8: Conclude position of $C'$ and $F$

Since $C'$ and $F$ are the same distance along the same ray from $D$, $C'$ and $F$ coincide.

Step9: Prove triangle congruence

A rigid motion takes $A$ to $D$, $B$ to $E$, and $C$ to $F$, so $\triangle ABC\cong\triangle DEF$.

Answer:

$\triangle ABC$ is congruent to $\triangle DEF$ as shown by the series of rigid - motion and congruence - property applications.