Question

given: \\( \overline{ad} \cong \overline{bc} \\) and \\( \overline{ad} \parallel \overline{bc} \\) prove: abcd is a parallelogram.

Explanation:

Step1: Connect \(AC\)

Draw diagonal \(AC\).

Step2: Prove \(\triangle ADC\cong\triangle CBA\)

Since \(AD\parallel BC\), then \(\angle DAC=\angle BCA\) (alternate - interior angles).
Given \(AD = BC\) (because \(\overline{AD}\cong\overline{BC}\)), and \(AC = CA\) (common side).
By the SAS (Side - Angle - Side) congruence criterion, \(\triangle ADC\cong\triangle CBA\).

Step3: Get \(AB = CD\)

From \(\triangle ADC\cong\triangle CBA\), we have \(AB = CD\) (corresponding parts of congruent triangles).

Step4: Use the parallelogram definition

A quadrilateral with both pairs of opposite sides parallel and equal is a parallelogram.
We know \(AD\parallel BC\) and \(AD = BC\), also \(AB = CD\).
Another way: Since \(\triangle ADC\cong\triangle CBA\), then \(\angle ACD=\angle CAB\), so \(AB\parallel CD\) (alternate - interior angles).
Since \(AD\parallel BC\) and \(AB\parallel CD\), \(ABCD\) is a parallelogram.

Answer:

Since \(AD\parallel BC\) and \(AD = BC\) (given \(\overline{AD}\cong\overline{BC}\)), and by proving \(\triangle ADC\cong\triangle CBA\) (using \(AD = BC\), \(\angle DAC=\angle BCA\) (from \(AD\parallel BC\)), \(AC = CA\) (common side) via SAS), we get \(AB = CD\) and \(AB\parallel CD\) (from congruent triangles and alternate - interior angles). A quadrilateral with both pairs of opposite sides parallel (\(AD\parallel BC\) and \(AB\parallel CD\)) is a parallelogram. So \(ABCD\) is a parallelogram.