Question

given: abcd is a parallelogram. prove: \\(\overline{ab} \cong \overline{cd}\\) and \\(\overline{bc} \cong \overline{da}\\) diagram of parallelogram abcd with vertices a, b, c, d; two - column proof template with statements and reasons, and drag - and - drop options: angles, segments, triangles, statements, reasons; symbols \\(\cong\\), \\(\parallel\\); segments \\(\overline{da}\\), \\(\overline{ab}\\), \\(\overline{ac}\\), \\(\overline{bc}\\), \\(\overline{cd}\\)

Explanation:

Step1: Recall parallelogram properties

In a parallelogram \(ABCD\), opposite sides are equal and parallel. So, \(AB \parallel CD\) and \(AB = CD\); \(BC \parallel DA\) and \(BC = DA\).

Step2: Match with given proof

We need to prove \(AB = CD\) and \(BC = DA\). From parallelogram side properties, \(AB\) and \(CD\) are opposite sides, \(BC\) and \(DA\) are opposite sides. So the segments to use are \(AB\), \(CD\), \(BC\), \(DA\) with equality and parallelism as per parallelogram definition.

Answer:

To prove \( \overline{AB} \cong \overline{CD} \) and \( \overline{BC} \cong \overline{DA} \) in parallelogram \(ABCD\):

• By definition of a parallelogram, opposite sides are parallel and congruent. Thus, \( \overline{AB} \parallel \overline{CD} \) and \( \overline{AB} \cong \overline{CD} \); \( \overline{BC} \parallel \overline{DA} \) and \( \overline{BC} \cong \overline{DA} \).

(For the drag - and - drop: Use \( \overline{AB} \cong \overline{CD} \) (from "Segments" with \( \cong \)) and \( \overline{BC} \cong \overline{DA} \) (from "Segments" with \( \cong \)), with the reason being "Opposite sides of a parallelogram are congruent".)