Question

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

Answer 1

Answer:

Step-by-step explanation:

Let a rectangle named ABCD, having diagonal AC and BD respectively.

Let the meeting point of two diagonal be O.

As we know that diagonal of a parallelogram bisect each other at 90 degree.

Therefore, angle BOC =AngleCOD=angle DOA=angleAOB

Now, in triangle AOB,

OB=OC. (Diagonals of a parallelogram bisect each other )

So,triangle AOB is an isosceles triangle.

As base angles are equal and one angle of an isosceles triangle is 90 degree.

Other two angle in all four triangle formed is 45 degree

Therefore,

Angle A=90 degree

Angle B=90 degree

Angle C=90 degree

Angle D=90 degree

Therefore, ABCD is an triangle.

Hence proved.

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)