Chapter 5.4, Problem 35WE

To determine

ToProve: if the diagonals of parallelograms are congruent, then the parallelogram is a rectangle.

Explanation of Solution

Given

The diagonals of a parallelogram are congruent.

Calculation:

Figure is made having diagonals AC and BD.

  

Here, $\overrightarrow{AC}\cong \overrightarrow{BD}$ .

In $\Delta ABC\text{ and }\Delta ABD$ .

  $\begin{array}{l}AB=AB\therefore \left(\text{Common}\right)\\ AC=BD\therefore \left(\text{Given}\right)\\ BC=AD\therefore \left(\text{Sides of a }▱\right)\end{array}$

So, $\Delta ABC\cong \Delta BAD$ by SSS postulate.

Therefore, by CPCT $\angle ABC\cong \angle BAD$ .

Using property of co-interior angles in a parallelogram.

  $\begin{array}{l}\angle ABC+\angle BAD=180\\ 2\angle ABC=180\therefore \left(\angle ABC\cong \angle BAD\right)\\ \angle ABC=90\end{array}$

Therefore, parallelogram ABCD is a rectangle.