In the parallelogram \(ABCD\), if the diagonals \(AC\) and \(BD\) are equal, then prove that \(ABCD\) is a rectangle.
Proof:
Let \(ABCD\) be a parallelogram.
In \(∆ABC\) and \(∆BCD\),
\(AC = BD\) [Given]
In \(∆ABC\) and \(∆BCD\),
\(AC = BD\) [Given]
\(AC = BD\) []
\(BC = CB\) [Common side]
Therefore, \(∆ABC ≅ ∆BCD\) []
So, \(∠ABC = ∠BCD\) [By C.P.C.T.] ----(1)
Now, \(AB || CD\) and \(BC\) is a transversal.
Now, \(AB || CD\) and \(BC\) is a transversal.
Therefore, \(∠ABC + ∠BCD = 180^°\) [Co-interior angles of parallelogram] ----- (2)
From (1) and (2), we have \(∠ABC = ∠BCD =\)\(^°\)
That is, \(ABCD\) is a parallelogram having an angle to \(90^°\).
Hence, \(ABCD\) is a rectangle.