If the diagonals of a parallelogram …

Given: The diagonals of a parallelogram are equal.

To prove: Parallelogram is a rectangle.
Proof : In {tex}\triangle{/tex}ACB and {tex}\triangle{/tex}BDA,
AC = BD . . . [Given]
AB = BA . . . [Common]
BC = AD . . . [Opposite sides of parallelogram]
{tex}\therefore{/tex} {tex}\triangle{/tex}ACB {tex}\cong{/tex}{tex}\triangle{/tex}BDA . . .[By SSS property]
{tex}\therefore{/tex} {tex}\angle{/tex}ABC = {tex}\angle{/tex}BAD . . . [c.p.c.t.] . . . .(1)
As AD || BC . . . [Opposite sides of parallelogram]
transversal AB intersects them.
{tex}\therefore{/tex} {tex}\angle{/tex}BAD + {tex}\angle{/tex}ABC = 180^o . . . [Sum of interior angle on the same side of a transversal] . . . .(2)
{tex}\angle{/tex}BAD = {tex}\angle{/tex}ABC = 90^o . . . [From (1) and (2)]
{tex}\therefore{/tex} {tex}\angle{/tex}A = 90^{o
{tex}\therefore{/tex}} Parallelogram ABCD is a rectangle.