Question

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

Answer 1

(i) It is given that ABCD is a rectangle.

∠A =∠C

$⇒ \frac{1}{2}∠A=\frac{1}{2}∠C$

$⇒∠DCA=∠DCA$ (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 ∆BCD,

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

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

However, ∠CDB = ∠BD (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.