Chapter 5.4, Problem 5CE

To determine

To prove: if one angle of a parallelogram is a right angle then it is a rectangle.

Explanation of Solution

Proof:

  

Assume that $\angle A=90°$

ABCD is a parallelogram.

  $AD\parallel BC$ (opposite side of a parallelogram are parallel)

  $\angle A+\angle B=180°$ (Adjacent angles of a parallelogram are supplementary)

  $\begin{array}{l}90°+\angle B=180°\\ \Rightarrow \angle B=180°-90°=90°\end{array}$

  $AB\parallel CD$ (opposite side of a parallelogram are parallel)

  $\begin{array}{l}\angle A+\angle D=180°\\ \Rightarrow 90°+\angle D=180°\\ \Rightarrow \angle D=180°-90°=90°\end{array}$

(Adjacent angles of a parallelogram are supplementary)

  $AD\parallel BC$ (opposite side of a parallelogram are parallel)

  $\begin{array}{l}\angle D+\angle C=180°\\ \Rightarrow 90°+\angle C=180°\\ \Rightarrow \angle C=180°-90°=90°\end{array}$

(Adjacent angles of a parallelogram are supplementary)

So, ABCD is a parallelogram such that $\angle A=\angle B=\angle C=\angle D=90°$

Therefore,ABCD is a rectangle.

(A parallelogram is which each angle is equal to $90°$ is a rectangle)