Question

question
given: abcd is a parallelogram, e is the midpoint of \\(\overline{ab}\\) and f is the midpoint of \\(\overline{dc}\\).
prove: \\(\overline{de} \cong \overline{fb}\\).

step | statement | reason
1 | abcd is a parallelogram; e is the midpoint of \\(\overline{ab}\\); f is the midpoint of \\(\overline{dc}\\) | given
2 | \\(\overline{ad} \cong \overline{cb}\\) | select a reason...
image of parallelogram abcd with e on ab, f on dc, and segments de, fb drawn

Explanation:

Step1: Recall properties of parallelograms

In a parallelogram, opposite sides are congruent. For parallelogram \(ABCD\), sides \(AD\) and \(CB\) are opposite sides.
So, the reason for \(\overline{AD} \cong \overline{CB}\) is "Opposite sides of a parallelogram are congruent".

Step2: Analyze midpoints and side lengths

Since \(E\) is the midpoint of \(\overline{AB}\) and \(F\) is the midpoint of \(\overline{DC}\), and in a parallelogram \(AB = DC\) (opposite sides of a parallelogram are congruent), we have \(AE = EB=\frac{1}{2}AB\) and \(DF = FC=\frac{1}{2}DC\). Thus, \(EB = DF\) (because \(AB = DC\) implies \(\frac{1}{2}AB=\frac{1}{2}DC\)).

Step3: Check for congruent triangles (optional for this step, but to prove \(DE\cong FB\))

We can show that \(\triangle ADE \cong \triangle CBF\) (or \(\triangle DFB \cong \triangle BED\)) using SAS (Side - Angle - Side) congruence criterion. But for the given step 2, the key is the property of parallelograms for \(\overline{AD} \cong \overline{CB}\).

Answer:

The reason for \(\overline{AD} \cong \overline{CB}\) is "Opposite sides of a parallelogram are congruent".