Question

given: abcd is a parallelogram. prove: ∠a≅∠c and ∠b≅∠d by the definition of a ☐, ad∥bc and ab∥dc. using, ad as a transversal, ∠a and ∠ are same - side interior angles, so they are. using side as a transversal, ∠b and ∠c are same - side interior angles, so they are. using ab as a transversal, ∠a and ∠ interior angles, so they are supplementary. therefore, ∠a is congruent to ∠c because they are supplements of the same angle. similarly, ∠b is congruent to ∠

Explanation:

Step1: Recall properties of parallel - lines

In a parallelogram \(ABCD\) with \(AD\parallel BC\) and \(AB\parallel DC\). When \(AD\) is a transversal for \(AB\) and \(DC\), \(\angle A\) and \(\angle D\) are same - side interior angles, so they are supplementary (\(\angle A+\angle D = 180^{\circ}\)).

Step2: Consider another transversal

When \(BC\) is a transversal for \(AB\) and \(DC\), \(\angle B\) and \(\angle C\) are same - side interior angles, so they are supplementary (\(\angle B+\angle C = 180^{\circ}\)).

Step3: Use transitive property

Since \(\angle A+\angle D = 180^{\circ}\) and \(\angle B+\angle C = 180^{\circ}\), and also when \(AB\) is a transversal for \(AD\) and \(BC\), \(\angle A+\angle B = 180^{\circ}\) and when \(DC\) is a transversal for \(AD\) and \(BC\), \(\angle C+\angle D = 180^{\circ}\). We know that \(\angle A\) and \(\angle C\) are supplements of the same angle (\(\angle D\) and \(\angle B\) respectively). So \(\angle A\cong\angle C\). Similarly, \(\angle B\) and \(\angle D\) are supplements of the same angle (\(\angle A\) and \(\angle C\) respectively), so \(\angle B\cong\angle D\).

Answer:

In the blanks: When using \(AD\) as a transversal, \(\angle A\) and \(\angle D\) are same - side interior angles. Using side \(BC\) as a transversal, \(\angle B\) and \(\angle C\) are same - side interior angles. \(\angle B\) is congruent to \(\angle D\).