Question

step | statement | reason
1 | abcd is a parallelogram, \\(\overline{ce} \cong \overline{fa}\\) | given
2 | \\(\overline{bc} \cong \overline{ad}\\) | opposite sides of a parallelogram / rhombus / rectangle / square are congruent
3 | \\(\overline{be} \cong \overline{fd}\\) | congruent segments added to congruent segments form congruent segments
4 | \\(\overline{bc} \parallel \overline{ad}\\) | opposite sides of a parallelogram / rhombus / rectangle / square are parallel
5 | \\(\overline{be} \parallel \overline{fd}\\) | opposite sides of a parallelogram / rhombus / rectangle / square are parallel
6 | fbed is a parallelogram | fbed is a parallelogram because it is a quadrilateral with select a reason...

diagram: a parallelogram abcd with extensions to form fbed, showing segment markings and parallel arrows

Explanation:

Step1: Recall parallelogram criteria

A quadrilateral is a parallelogram if both pairs of opposite sides are congruent and parallel, or one pair of opposite sides is both congruent and parallel. From steps 3 and 5, we have \(\overline{BE}\cong\overline{FD}\) (congruent opposite sides) and \(\overline{BE}\parallel\overline{FD}\) (parallel opposite sides).

Step2: Apply the parallelogram definition

Since \(FBED\) is a quadrilateral with one pair of opposite sides (\(\overline{BE}\) and \(\overline{FD}\)) that are both congruent and parallel, or using the criteria: "If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram" (we already have \(\overline{BE}\cong\overline{FD}\) and from the parallelogram \(ABCD\) and the congruent segments addition, we can also infer the other pair, but more directly, from steps 3 (congruent) and 5 (parallel) for \(\overline{BE}\) and \(\overline{FD}\), and similarly we can check the other pair, but the key reason here is that in step 3 we have \(\overline{BE}\cong\overline{FD}\) (opposite sides congruent) and step 5 \(\overline{BE}\parallel\overline{FD}\) (opposite sides parallel), so the reason is "Both pairs of opposite sides are congruent and parallel" or more precisely, from the previous steps, since we have \(\overline{BE}\cong\overline{FD}\) (step 3) and \(\overline{BE}\parallel\overline{FD}\) (step 5), and also we can see \(\overline{BF}\) and \(\overline{ED}\) would also be congruent and parallel, but the standard reason is: "If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram" (or "If one pair of opposite sides of a quadrilateral is both congruent and parallel, then the quadrilateral is a parallelogram"). Given the previous steps, since we have \(\overline{BE}\cong\overline{FD}\) (congruent) and \(\overline{BE}\parallel\overline{FD}\) (parallel), the reason is that a quadrilateral with one pair of opposite sides both congruent and parallel is a parallelogram, or using the two - part: both pairs of opposite sides are congruent (from step 3: \(\overline{BE}\cong\overline{FD}\), and we can also show \(\overline{BF}\cong\overline{ED}\) similarly, but from the given steps, step 3 gives congruent opposite sides and step 5 gives parallel opposite sides, so the reason is "Both pairs of opposite sides are congruent (and parallel)" or the specific theorem: "If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram" (we have \(\overline{BE}\cong\overline{FD}\) and we can derive \(\overline{BF}\cong\overline{ED}\) as well, but from the steps, step 3 is \(\overline{BE}\cong\overline{FD}\) (opposite sides congruent) and step 5 is \(\overline{BE}\parallel\overline{FD}\) (opposite sides parallel), so the reason is "Both pairs of opposite sides are congruent and parallel" or the theorem: "A quadrilateral with both pairs of opposite sides congruent is a parallelogram" (since we have \(\overline{BE}\cong\overline{FD}\) and we can get \(\overline{BF}\cong\overline{ED}\) by similar logic as step 3 for the other pair).

Answer:

The reason is "Both pairs of opposite sides are congruent (or: If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram)" (more precisely, from steps 3 (\(\overline{BE}\cong\overline{FD}\)) and the other pair can be shown congruent too, but the key is that we have one pair of opposite sides congruent and parallel, or both pairs congruent. The standard reason here is "If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram" (since we have \(\overline{BE}\cong\overline{FD}\) from step 3, and we can derive \(\overline{BF}\cong\overline{ED}\) as follows: from \(ABCD\) being a parallelogram, \(\overline{AB}\cong\overline{CD}\), and with \(\overline{CE}\cong\overline{FA}\), we can show \(\overline{BF}=\overline{BA}+\overline{AF}\) and \(\overline{ED}=\overline{DC}+\overline{CE}\), and since \(\overline{BA}\cong\overline{DC}\) (opposite sides of parallelogram \(ABCD\)) and \(\overline{AF}\cong\overline{CE}\) (given \(\overline{CE}\cong\overline{FA}\)), so \(\overline{BF}\cong\overline{ED}\) by congruent segments addition. Thus both pairs of opposite sides (\(\overline{BE}\cong\overline{FD}\) and \(\overline{BF}\cong\overline{ED}\)) are congruent, so the quadrilateral \(FBED\) is a parallelogram by the theorem "If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram".