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

Let ABCD be a parallelogram with equal diagonals.

From figure,

In ∆ ABC and ∆ DCB,

⇒ AB = DC (Opposite sides of a parallelogram are equal)

⇒ BC = BC (Common side)

⇒ AC = DB (Diagonals of parallelogram are equal)

∴ ∆ ABC ≅ ∆ DCB (By S.S.S. Congruence rule)

We know that,

Corresponding parts of congruent triangles are equal.

⇒ ∠ABC = ∠DCB (By C.P.C.T.) ……(1)

We know that,

Sum of co-interior angles equal to 180°.

⇒ ∠ABC + ∠DCB = 180° (AB || CD)

⇒ ∠ABC + ∠ABC = 180° [From equation (1)]

⇒ 2∠ABC = 180°

⇒ ∠ABC = $\dfrac{180°}{2}$

⇒ ∠ABC = ∠B = 90°

⇒ ∠DCB = ∠C = 90°

Since,

Opposite angles of parallelogram are equal.

∴ ∠D = ∠B = 90° and ∠A = ∠C = 90°.

∴ ∠B = ∠D = ∠C = ∠A = 90°.

Since, opposite sides of ABCD are equal in length and each interior angle equals to 90°.

Hence, proved that if the diagonals of a parallelogram are equal, then it is a rectangle.