Chapter 5.6, Problem 13PSB

To determine

To prove:A quadrilateralABCD is a parallelogram.

Explanation of Solution

Given information:

InquadrilateralABCD,

  $\begin{array}{l}\angle A={\left(x^5.x^2\right)}^{\circ },\\ \angle B={\left(x^2-25\right)}^{\circ },\\ \angle C={\left(x^7\right)}^{\circ },\\ \angle D={\left((x-5)(x+5)\right)}^{\circ }.\end{array}$

Proof:

  

  $\begin{array}{l}\angle A={\left(x^5.x^2\right)}^{\circ }={\left(x^7\right)}^{\circ },\\ \angle C={\left(x^7\right)}^{\circ }.\\ \angle A\cong \angle C\end{array}$

  $\begin{array}{l}\angle B={\left(x^2-25\right)}^{\circ },\\ \angle D={\left((x-5)(x+5)\right)}^{\circ }={\left(x^2-25\right)}^{\circ }.\\ \angle B\cong \angle D\end{array}$

Because angle B and D are congruent and form straight angles with their adjacent angles, the adjacent angles are congruent. Thus, the quadrilateralABCD is parallelogram because the two pairs of opposite angles are congruent.