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.
(i) It is given that ABCD is a rectangle.
∴ ∠A = ∠C
⇒ 1/2∠A = 1/2∠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.
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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