Question

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

Answer 1

Hey Yogi., here is your answer..........❗❗❤☺⤵


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#,Let ABCD be a parallelogram. To show that ABCD is a rectangle, we have to prove that one of its interior angles is 90º.

In ΔABC and ΔDCB,

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

BC = BC (Common)

AC = DB (Given)

∴ ΔABC ≅ ΔDCB (By SSS Congruence rule)

⇒ ∠ABC = ∠DCB

It is known that the sum of the measures of angles on the same side of transversal is 180º.

∠ABC + ∠DCB = 180º (AB || CD)

⇒ ∠ABC + ∠ABC = 180º

⇒ 2∠ABC = 180º

⇒ ∠ABC = 90º

Since ABCD is a parallelogram and one of its interior angles is 90º, ABCD is a rectangle......❗❗☺❤

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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)