ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C Show that:

(i) ABCD is a square

(ii) Diagonal BD bisects ∠B as well as ∠D

It is given that ABCD is a rectangle

∠A = ∠C

∠A = ∠C

∠DAC =∠ DCA (AC bisects A and C)

CD = DA (Sides opposite to equal angles are also equal)

However,

DA = BC and AB = CD (Opposite sides of a rectangle are equal)

AB = BC = CD = DA

ABCD is a rectangle and all of its sides are equal.

Hence, ABCD is a square

(ii) Let us join BD

In ΔBCD,

BC = CD (Sides of a square are equal to each other)

∠CDB = ∠CBD (Angles opposite to equal sides are equal)

However,

∠CDB = ∠ABD (Alternate interior angles for AB || CD)

∠CBD = ∠ABD

BD bisects B

Also,

∠CBD = ∠ADB (Alternate interior angles for BC || AD)

∠CDB = ∠ABD

BD bisects D