(i) Given: ABCD is a rectangle ∠1 = ∠2 and ∠3 = ∠4.
To prove ABCD is a square.
Solution: ∠1 = ∠4 …(1) [∵ Alternate angle]
∠3 = ∠4 …(2) [∵ Given]
Hence, ∠1 = ∠3 …(3) [∵ From (1) and (2)]
In ΔADC,
∠1 = ∠3 [∵ From (3)]
DC = AD [∵ In a triangle, side opposite to equal angle are equal]
A rectangle, whose adjacent sides are equal, is a square.
Hence, ABCD is a square.
(ii) To prove: Diagonal BD bisects angle B as well as angle D.
Solution: ∠5 = ∠8 …(4) [ Alternate angle]
In ΔADB,
AB = AD [∵ ABCD is a square]
∠7 = ∠5 …(5) [∵ Angles opposite to equal sides are equal]
Hence, ∠7 = ∠8 …(6) [∵ From (4) and (5)]
and ∠7 = ∠6 …(7) [∵ Alternate angles]
Hence, ∠5 = ∠6 …(8) [∵ From (5) and (7)]
Hence, from (6) and (8), diagonal BD Bisects angle B as well as D.