Question

if the diagonal of a parallelogram are equal then show that is a rectangle​

Answer 1

hyea!!

HETE IS YOUR ANSWER.....

Q)- If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Answer:

Let ABCD be a parallelogram. To show that ABCD is a rectangle, we have to prove that one of

its interior angles is 900.

In ΔABC and ΔDCB,

AB = DC (Opposite sides of a parallelogram are equal)

BC = BC (Common)

AC = DB (Given)

By SSS congruence rule,

ΔABC ≅ ΔDCB

So, ∠ABC = ∠DCB

It is known that the sum of measures of angles on the same side of traversal is 1800

∠ABC + ∠DCB = 1800 [AB || CD]

=> ∠ABC + ∠ABC = 1800

=> 2∠ABC = 1800

=> ∠ABC = 900

Since ABCD is a parallelogram and one of its interior angles is 900, 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)