Chapter 5.6, Problem 20PSC

To determine

To prove:ABCDis a parallelogram.

Explanation of Solution

Given information:

Two parallel lines with a quadrilateral ABCDform congruent angles.

Proof:

  

  $\begin{array}{l}\angle 1\cong \angle 4\\ \angle 2\cong \angle 3\\ \angle 1+\angle ABC+\angle 2=\angle 4+\angle ADC+\angle 3={180}^{\circ }\end{array}$

If you draw BD, then $\angle 1+\angle ABD\cong \angle 4+\angle BDC$ because they are alternate interior angles between parallel lines.

  $\begin{array}{l}\angle 1\cong \angle 4\\ \angle ABD\cong \angle BDC\\ \angle 2+\angle CBD=\angle 3+\angle ADB\end{array}$

If lines are parallel, then alternate interior angles are congruent.

  $\begin{array}{l}\angle 2\cong \angle 3\\ \angle CBD\cong \angle ADB\\ \overline{BD}\cong \overline{BD}\end{array}$

Two triangles are congruent by ASA congruence rule.

  $\Delta ABD\cong \Delta CDB$

Corresponding parts of congruent triangles are congruent.

  $\angle A\cong \angle C$

As opposite angles are congruent,ABCD is a parallelogram.