Question

question
given: abcd is a parallelogram and b is the midpoint of \\(\overline{af}\\).
prove: \\(\overline{df}\\) bisects \\(\overline{bc}\\).
step\tstatement\treason
1\tabcd is a parallelogram
b is the midpoint of \\(\overline{af}\\)\tgiven
2\t\\(\overline{ab} \cong \overline{bf}\\)\ta midpoint divides a segment into two congruent segments
3\t\\(\overline{ab} \cong \overline{dc}\\)\topposite sides of a parallelogram / rhombus / rectangle / square are congruent
4\t\\(\overline{bf} \cong \overline{dc}\\)\ttransitive property
5\t\\(\overline{dc} \parallel \overline{ab}\\)\tselect a reason...

Explanation:

Step 1: Recall Parallelogram Property

In a parallelogram, opposite sides are parallel. Since \(ABCD\) is a parallelogram, \(AB\parallel DC\) (by the definition of a parallelogram: opposite sides of a parallelogram are parallel).
So the reason for \(\overline{DC}\parallel\overline{AB}\) is "Opposite sides of a parallelogram are parallel".

Step 2: Prove Triangles Congruent (Optional for Completion)

Let's continue to fully prove \(DF\) bisects \(BC\). Let \(E\) be the intersection of \(DF\) and \(BC\).

• From Step 4, \(\overline{BF}\cong\overline{DC}\), and from Step 5, \(\overline{DC}\parallel\overline{AB}\) (so \(\overline{DC}\parallel\overline{BF}\) as \(AB\) and \(BF\) are on the same line).
• \(\angle CED\cong\angle BEF\) (vertical angles are congruent).
• \(\angle CDE\cong\angle BFE\) (alternate interior angles, since \(DC\parallel BF\)).
• By AAS (Angle - Angle - Side) congruence criterion, \(\triangle DCE\cong\triangle FBE\).
• Then \(\overline{CE}\cong\overline{BE}\) (corresponding parts of congruent triangles are congruent), which means \(DF\) bisects \(BC\).

Answer:

The reason for \(\overline{DC}\parallel\overline{AB}\) is "Opposite sides of a parallelogram are parallel". And by proving \(\triangle DCE\cong\triangle FBE\) (using AAS with the established parallelism and congruence), we conclude \(DF\) bisects \(BC\) as \(\overline{CE}\cong\overline{BE}\).