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.

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Answer 1

$\huge\mathbb{SOLUTION:-}$

Given:-

• ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C.

Explanation:-

(i) ∠DAC = ∠DCA (AC bisects ∠A as well as ∠C)

⇒ AD = CD (Sides opposite to equal angles of a triangle are equal)

also, CD = AB (Opposite sides of a rectangle)

AB = BC = CD = AD

Thus, ABCD is a square.

(ii) In ΔBCD,BC = CD

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

also, ∠CDB = ∠ABD (Alternate interior angles)

⇒ ∠CBD = ∠ABD

Thus, BD bisects ∠B

Now,

∠CBD = ∠ADB

⇒ ∠CDB = ∠ADB

Thus, BD bisects ∠D

Answer 2

Step-by-step explanation:

Given:-

ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C.

Explanation:-

(i) ∠DAC = ∠DCA (AC bisects ∠A as well as ∠C)

⇒ AD = CD (Sides opposite to equal angles of a triangle are equal)

also, CD = AB (Opposite sides of a rectangle)

AB = BC = CD = AD

Thus, ABCD is a