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 diagram with points f, a, d on the bottom side and b, c, e on the top side, with markings for congruent segments

Explanation:

Step1: Analyze known congruences

We know \( \overline{BC} \cong \overline{AD} \) (from step 2, opposite sides of parallelogram) and \( \overline{CE} \cong \overline{FA} \) (given).

Step2: Use segment addition

For \( \overline{BE} \), it is \( \overline{BC} + \overline{CE} \), and for \( \overline{FD} \), it is \( \overline{AD} + \overline{FA} \). Since \( \overline{BC} \cong \overline{AD} \) and \( \overline{CE} \cong \overline{FA} \), by the Segment Addition Postulate and the property of congruent segments (if \( a \cong b \) and \( c \cong d \), then \( a + c \cong b + d \)), we get \( \overline{BE} \cong \overline{FD} \). The reason is: If two congruent segments are added to two other congruent segments, the resulting segments are congruent (or more formally, using the addition of congruent segments: If \( \overline{BC} \cong \overline{AD} \) and \( \overline{CE} \cong \overline{FA} \), then \( \overline{BC}+\overline{CE} \cong \overline{AD}+\overline{FA} \), so \( \overline{BE} \cong \overline{FD} \)).

Answer:

The reason for \( \overline{BE} \cong \overline{FD} \) is: If two congruent segments are added to two other congruent segments, the resulting segments are congruent (or "Addition of congruent segments: If \( \overline{BC} \cong \overline{AD} \) and \( \overline{CE} \cong \overline{FA} \), then \( \overline{BC}+\overline{CE} \cong \overline{AD}+\overline{FA} \), so \( \overline{BE} \cong \overline{FD} \)").