Question

if the diagonals of a parallelogram are equal then show that it is a rectangle

Answer 1

Let ABCD be the parallelogram.
Given,AC=BD
We know,AB=CD(opposite sides of parallelogram are equal)
BC=BC
According to SSS congruence rule 
ΔABC congruent to ΔDCB
So, angle ABC= angle DCB (CPCT)
angle ABC+angle DCB=180(co-interior angles)
2*(angle ABC)=180(ABC+DCB)
angle ABC=90
So ABCD is a rectangle.

Answer 2

Given: ABCD is a parallelogram and AC = BD

To prove: ABCD is a rectangle

Proof:  In  Δ ACB and ΔDCB

AB = DC _____ Opposite sides of parallelogram are equal

BC = BC _____ Common side

AC = DB _____ Given

Therefore,

Δ ACB ≅ ΔDCB by S.S.S test

Angle ABC = Angle DCB ______ C.A.C.T

Now,

AB ║ DC _______ Opposite sides of parallelogram are parallel

Therefore,

Angle B + Angle C = 180 degree (Interior angles are supplementary)

Angle B + Angle B = 180

2 Angle B  = 180 degree

Angle B = 90 degree

Similarly, we can prove that, Angle A = 90 degree, Angle C = 90 degree and Angle D = 90 degree.

Therefore, ABCD is a rectangle.

(Refer to the attachment for the figure)