Q. 84.1( 435 Votes )

# 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 :

**Given: ABCD is a rectangle, **

**∠A = ∠C**

** ∠A = ∠CTo Prove: ABCD is a squareProof:**

∠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

Hence, Proved.**(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

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