In a parallelogram \(PQRS\), the diagonals
\(PR\) and
\(QS\) are equal. Prove that
\(
\angle PQR = 90°\). Hence show that
\(PQRS\) is a rectangle.
Proof:
Let \(PQRS\) be a parallelogram.
In \(∆PQR\) and \(∆QRS\),
\(PR = QS\) [Given]
In \(∆PQR\) and \(∆QRS\),
\(PR = QS\) [Given]
\(PQ = RS\) []
\(QR = RQ\) [Common side]
Therefore, \(∆PQR ≅ ∆QRS\) []
So, \(∠PQR = ∠QRS\) [By C.P.C.T.] ----(1)
Now, \(PQ || RS\) and \(QR\) is a transversal.
Now, \(PQ || RS\) and \(QR\) is a transversal.
Therefore, \(∠PQR + ∠QRS = 180^°\) [Co-interior angles of parallelogram] ----- (2)
From (1) and (2), we have \(∠PQR = ∠QRS =\)\(^°\)
That is, \(PQRS\) is a parallelogram having an angle to \(90^°\).
Hence, \(PQRS\) is a rectangle.