Question

Question 2
If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Answer 1

Let ABCD be a parallelogram, To show that ABCD is a rectangle, we have to prove that one of its interior angles is ${90}^{\circ }$.
In $\Delta$ ABC and $\Delta$BAD,
BC=AD ( Opposite sides of a parallelogram are equal)
AB=AB ( common)
AC=DB (Given)
$\therefore \Delta ABC\cong BAD$ ( By SSS congruence rule)
$\angle ABC=\angle BAD$………(1)
It is known that the sum of the measures of angles on the same side of transversal is ${180}^{\circ }$
$\Rightarrow \angle ABC+\angle BAD={180}^{\circ }$ ( AB = CD)
$\Rightarrow \angle ABC+\angle ABC={180}^{\circ }$ (From (1))
$\Rightarrow 2\angle ABC={180}^{\circ }$
$\Rightarrow \angle ABC={90}^{\circ }$
Since ABCD is a parallelogram and one of its interior angles is ${90}^{\circ }$, ABCD is a rectangle.