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

Given: Parallelogram is ABCD whose diagonals AC and BD are equal.



\(\Rightarrow \) AC = BD

To prove: ABCD is a rectangle.

Proof:
In \(\triangle{ABC}\) and \(\triangle{DCB}\), we have,
AC = BD ...(Given)
AB = CD
(Opposite sides of parallelogram)
BC = CB ...(Common sides)
\(\therefore \) \(\triangle{ABC}\)\(\displaystyle \cong \)\(\triangle{DCB}\)
(By SSS rule)
\(\therefore \) \(\angle{ABC}\) = \(\angle{DCB}\) ...(i)(By CPCT)

But from figure, DC || AB and transversal CB intersects them

\(\angle{ABC}\) + \(\angle{DCB}\) = \(180^\circ\)
(interior angles on same side of transversal)
\(\angle{ABC}\) + \(\angle{ABC}\) = \(180^\circ\)
(from(i))
\(\Rightarrow \) 2 \(\angle{ABC}\) = \(180^\circ\)
\(\Rightarrow \) \(\angle{ABC}\) = \(90^\circ\)
Also, \(\angle{DCB}\) = \(90^\circ\)

Thus, we can say that, ABCD is a parallelogram and some angles are \(90^\circ\).

Hence it is proved that, ABCD is a rectangle.