Chapter 6.3, Problem 30PPS

To determine

To prove:

The paragraph proof for the theorem 6.10:

  $\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$ is a parallelogram

Explanation of Solution

Given information:

  $\angle \mathrm{A}\cong \angle \mathrm{C},\angle \mathrm{B}\cong \angle \mathrm{D}$

Proof:

Draw $\overline{\mathrm{A}\mathrm{C}}$ to form two triangles. The sum of the angles of one triangle is $180°$ , so the sum of the angles for two triangles is $180°$ . So, $\mathrm{m}\angle \mathrm{A}+\mathrm{m}\angle \mathrm{B}+\mathrm{m}\angle \mathrm{C}+\mathrm{m}\angle \mathrm{D}=360$ Because $\angle \mathrm{A}\cong \angle \mathrm{C},\angle \mathrm{B}\cong \angle \mathrm{D},\mathrm{m}\angle \mathrm{A}=\mathrm{m}\angle \mathrm{C},\mathrm{m}\angle \mathrm{B}=\mathrm{m}\angle \mathrm{D}$ . By substitution $\mathrm{m}\angle \mathrm{A}+\mathrm{m}\angle \mathrm{A}+\mathrm{m}\angle \mathrm{B}+\mathrm{m}\angle \mathrm{B}=360°$ . So $2\left(\mathrm{m}\angle \mathrm{A}\right)+2\left(\mathrm{m}\angle \mathrm{B}\right)=180°$ . Dividing each side by $2$ yields $\mathrm{m}\angle \mathrm{A}+\mathrm{m}\angle \mathrm{B}=180°$ . So, the consecutive angles are supplementary and $\overline{\mathrm{A}\mathrm{D}}\parallel \overline{\mathrm{B}\mathrm{C}}$ . Likewise, $2\left(\mathrm{m}\angle \mathrm{A}\right)+2\left(\mathrm{m}\angle \mathrm{D}\right)=0°$ or $\mathrm{m}\angle \mathrm{A}+\mathrm{m}\angle \mathrm{D}=180°$ . So, these consecutive angles are supplementary and $\overline{\mathrm{A}\mathrm{B}}\parallel \overline{\mathrm{D}\mathrm{C}}$ . Opposite sides are parallel, so $\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$ is a parallelogram.

Hence proved