Question

1. given: $bc \parallel ef$, $\overline{ab} \cong \overline{de}$, $\angle c \cong \angle f$

prove: $\triangle abc \cong \triangle def$

Explanation:

Step1: Identify corresponding angles

Since $\overline{BC} \parallel \overline{EF}$, the corresponding angles formed by transversal $\overline{AC}$ are congruent: $\angle ABC \cong \angle DEF$.

Step2: List given congruent parts

We know:

1. $\overline{AB} \cong \overline{DE}$ (Given)
2. $\angle ABC \cong \angle DEF$ (From Step1)
3. $\angle C \cong \angle F$ (Given)

Step3: Apply AAS congruence rule

By the Angle-Angle-Side (AAS) congruence postulate, if two angles and a non-included side of one triangle are congruent to the corresponding two angles and non-included side of another triangle, the triangles are congruent. Here, $\angle C \cong \angle F$, $\angle ABC \cong \angle DEF$, and $\overline{AB} \cong \overline{DE}$, so $\triangle ABC \cong \triangle DEF$.

Answer:

$\triangle ABC \cong \triangle DEF$ is proven by the AAS congruence postulate.