Question

a parrallogram having equal diagonal must be a rectangle prove it

Answer 1

Given- A ||gm ABCD with diagonals BD and AC intersect at O.

To prove- ABCD is a rectangle.

Proof- In ∆AOD and ∆ BOC:
AO = OC (diagonal bisect each other)
angle AOD = BOC (vertically opposite)
BO = OD (diagonal bisect each other)
so ∆AOD is congruent to ∆BOC (by SAS rule)

therefore OD = OC (by cpct)
and also angle ADO = BCO (by cpct)

in ∆OCD:since OD = OC
therefore angle ODC = OCD

now, adding these angled,
ADO+ODC = OCD+BCO
ADC=BCD

now, since AD||BC and DC is transversal,
so, angle ADC+BCD=180°(cointerior angles)
2 angle ADC=180° (ADC=BCD)
angle ADC=90°

so ABCD is a parallelogram where angle ADC is 90° therefore ABCD is a rectangle.

hope it helps you.