Question

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

Answer 1

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

In ΔPQR and ΔSRQ,

PQ = SR (Opposite sides of a parallelogram are equal)

QR = QR (Common)

PR = SQ (Given)

∴ ΔPQR ≅ ΔSRQ (By SSS Congruence rule)

⇒ ∠PQR = ∠SRQ

Since adjacent angles of a parallelogram are supplementary. (Consecutive interior angles)

∠PQR + ∠SRQ= 180º

⇒ ∠PQR + ∠PQR= 180º

⇒ 2∠PQR= 180º

⇒ ∠PQR = 90º

Since PQRS is a parallelogram and one of its interior angles is 90º, PQRS is a rectangle


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Answer 2

lets say, ABCD is a parallelogram


Given that the diagonals AC and BD of parallelogram ABCD are equal in length .

Consider triangles ABD and ACD.

AC = BD [Given]

AB = DC [opposite sides of a parallelogram]

AD = AD [Common side]

∴ ΔABD ≅ ΔDCA [SSS congruence criterion]

∠BAD = ∠CDA [CPCT]

∠BAD + ∠CDA = 180° [Adjacent angles of a parallelogram are supplementary.]

So, ∠BAD and ∠CDA are right angles as they are congruent and supplementary.

Therefore, parallelogram ABCD is a rectangle since a parallelogram with one right interior angle is a rectangle.




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