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# 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.