Question

given: abcd is a parallelogram and \\(\overline{ab} \cong \overline{ce}\\).
prove: \\(\angle a \cong \angle e\\).

step | statement | reason
1 | abcd is a parallelogram; \\(\overline{ab} \cong \overline{ce}\\) | given
2 | \\(\overline{cd} \cong \overline{ab}\\) | opposite sides of a parallelogram / rhombus / rectangle / square are congruent
3 | \\(\overline{cd} \cong \overline{ce}\\) | select a reason...

Explanation:

Step1: Identify the congruent segments

We know from Step 1 that $\overline{AB} \cong \overline{CE}$ (given) and from Step 2 that $\overline{CD} \cong \overline{AB}$ (opposite sides of a parallelogram are congruent).

Step2: Apply the Transitive Property of Congruence

The Transitive Property of Congruence states that if $a \cong b$ and $b \cong c$, then $a \cong c$. Here, let $a = \overline{CD}$, $b = \overline{AB}$, and $c = \overline{CE}$. Since $\overline{CD} \cong \overline{AB}$ and $\overline{AB} \cong \overline{CE}$, by the Transitive Property of Congruence, we can conclude that $\overline{CD} \cong \overline{CE}$.

Answer:

The reason for $\overline{CD} \cong \overline{CE}$ is the Transitive Property of Congruence (If two segments are congruent to the same segment, then they are congruent to each other).