Question

prove: \\(\overline{df}\\) bisects \\(\overline{bc}\\).
step | statement | reason
1 | \\(abcd\\) is a parallelogram; \\(b\\) is the midpoint of \\(\overline{af}\\) | given
2 | \\(\overline{ab} \cong \overline{bf}\\) | a midpoint divides a segment into two congruent segments
3 | \\(\overline{ab} \cong \overline{dc}\\) | opposite sides of a parallelogram / rhombus / rectangle / square are congruent
4 | \\(\overline{bf} \cong \overline{dc}\\) | transitive property
5 | \\(\overline{dc} \parallel \overline{ab}\\) | opposite sides of a parallelogram / rhombus / rectangle / square are parallel
6 | \\(\angle c \cong \angle ebf\\) | select a reason...
image of a parallelogram (abcd) with (f) on the extension of (ab) beyond (b), and (df) intersecting (bc) at (e)

Explanation:

Step 6: Identify the reason for ∠C ≅ ∠EBF

Since \( \overline{DC} \parallel \overline{AB} \) (from step 5, opposite sides of a parallelogram are parallel), and \( \overline{BC} \) is a transversal cutting these parallel lines, \( \angle C \) and \( \angle EBF \) are corresponding angles. By the Corresponding Angles Postulate, if two parallel lines are cut by a transversal, then corresponding angles are congruent. So the reason is "Corresponding Angles Postulate (If two parallel lines are cut by a transversal, corresponding angles are congruent)".

Step 7: Prove \( \triangle DCE \cong \triangle FBE \)

We have \( \overline{BF} \cong \overline{DC} \) (step 4), \( \angle C \cong \angle EBF \) (step 6), and \( \angle DEC \cong \angle BEF \) (vertical angles are congruent). By the AAS (Angle - Angle - Side) Congruence Theorem, \( \triangle DCE \cong \triangle FBE \).

Step 8: Conclude that \( \overline{CE} \cong \overline{BE} \)

Since \( \triangle DCE \cong \triangle FBE \), corresponding parts of congruent triangles are congruent (CPCTC). So \( \overline{CE} \cong \overline{BE} \), which means \( E \) is the midpoint of \( \overline{BC} \), so \( \overline{DF} \) bisects \( \overline{BC} \).

Answer:

For step 6, the reason is "Corresponding Angles Postulate (If two parallel lines are cut by a transversal, corresponding angles are congruent)". And by proving \( \triangle DCE \cong \triangle FBE \) (AAS) and using CPCTC, we conclude \( \overline{DF} \) bisects \( \overline{BC} \).