If ABCD is a rectangle in which the diagonal BD bisects ∠B, then show that ABCD is a square.

Since, BD bisects ∠B.

∴ ∠1 = ∠2

Since, ∠B = 90°.

∠1 = ∠2 = 45°

∠4 = ∠1 = 45° (Alternate angles are equal)

∠3 = ∠2 = 45° (Alternate angles are equal)

In △ABD,

AB = AD (As sides opposite to equal angles are equal) ………(i)

In △CBD,

BC = CD (As sides opposite to equal angles are equal) ……….(ii)

and AD = BC, AB = CD (As opposite sides of rectangle are equal) …….(iii)

From (i), (ii) and (iii) we get,

AB = BC = CD = AD.

Since, all sides are equal and alternate sides are perpendicular to each other.

Hence, proved that ABCD is a square.