Question

1. given: \\( \overline { d e } \parallel \overline { f g } \\), \\( \overline { d e } \cong \overline { f g } \\), \\( \angle d e f \cong \angle f g h \\) prove: \\( \angle d f e \cong \angle f h g \\)

Explanation:

Step1: Use parallel line property

Since \( \overline{DE}\parallel\overline{FG}\), by the alternate - interior angles theorem, \( \angle D = \angle GFH\).

Step2: Prove triangles congruent

We have \( \overline{DE}\cong\overline{FG}\), \( \angle DEF\cong\angle FGH\) (given), and \( \angle D=\angle GFH\) (from step 1). By the Angle - Side - Angle (\(ASA\)) congruence criterion, \( \triangle DEF\cong\triangle FGH\).

Step3: Use congruent triangle property

Because \( \triangle DEF\cong\triangle FGH\), corresponding parts of congruent triangles are congruent. So, \( \angle DFE\cong\angle FHG\).

Answer:

1. \( \angle D=\angle GFH\); Reason: Alternate - interior angles theorem (since \( \overline{DE}\parallel\overline{FG}\)).
2. \( \triangle DEF\cong\triangle FGH\); Reason: \(ASA\) ( \( \overline{DE}\cong\overline{FG}\), \( \angle DEF\cong\angle FGH\), \( \angle D=\angle GFH\)).
3. Reason: Corresponding parts of congruent triangles (\( \triangle DEF\cong\triangle FGH\)) are congruent.