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Answer :
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.