Question

given: abcd is a parallelogram and \\(\overline{ce} \cong \overline{fa}\\).
prove: fbed is a parallelogram.
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}\\) | select a reason...
(image of the parallelogram fbed with abcd as a parallelogram, points f, a, d on the base and b, c, e on the top, with segment markings indicating congruent sides)

Explanation:

Step1: Analyze known congruences

We know \( \overline{BC} \cong \overline{AD} \) (opposite sides of parallelogram \( ABCD \)) and \( \overline{CE} \cong \overline{FA} \) (given).

Step2: Use segment addition postulate

For \( \overline{BE} \), we have \( \overline{BE}=\overline{BC}+\overline{CE} \) (segment addition, since \( C \) is between \( B \) and \( E \)). For \( \overline{FD} \), we have \( \overline{FD}=\overline{FA}+\overline{AD} \) (segment addition, since \( A \) is between \( F \) and \( D \)).

Step3: Substitute congruent segments

Since \( \overline{BC} \cong \overline{AD} \) (so \( BC = AD \)) and \( \overline{CE} \cong \overline{FA} \) (so \( CE = FA \)), then \( BC + CE=AD + FA \). By substitution, \( BE = FD \), so \( \overline{BE} \cong \overline{FD} \) (definition of congruent segments: if lengths are equal, segments are congruent). The reason is "If two segments are each the sum of two congruent segments, then the sums are congruent" (or more formally, using segment addition and substitution with congruent segments).

Answer:

The reason for \( \overline{BE} \cong \overline{FD} \) is: If \( \overline{BC} \cong \overline{AD} \) and \( \overline{CE} \cong \overline{FA} \), then \( \overline{BC}+\overline{CE}\cong\overline{AD}+\overline{FA} \) (by the addition property of congruent segments, since \( \overline{BE}=\overline{BC}+\overline{CE} \) and \( \overline{FD}=\overline{FA}+\overline{AD} \) via segment addition postulate), so \( \overline{BE}\cong\overline{FD} \).