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Asked by Vishnu Last Modified
Answer 
Tasneem
(i) It is given that ABCD is a rectangle.
∴∠A = ∠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.
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