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

$\huge\star{\underline{\mathtt{\red{A}\pink{n}\green{s}\blue{w}\purple{e}\orange{r}}}}$

$\large\sf\underline\green{Given:-}$

$\sf\red{ABCD\:is\:a\:rectangle\:in\:which\:diagonal}$

$\sf\purple{AC\:bisects\:∠A\:as\:well\:as\:∠C}$

$\large\sf\underline\pink{Explanation:-}$

(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)

$\sf\green{AB\:=\:BC\:=\:CD\:=\:AD}$

$\sf\blue{Thus,\:ABCD\:is\:a\:square}$

(ii) In ΔBCD,BC = CD

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

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

⇒ ∠CBD = ∠ABD

$\sf\pink{Thus,\:BD\:bisects\:∠B}$

$\large\sf\underline\red{Now,}$

∠CBD = ∠ADB

⇒ ∠CDB = ∠ADB

$\sf\orange{Thus,\:BD\:bisects\:∠D}$

Answer 2

$\huge\star{\underline{\mathtt{\red{A}\pink{n}\green{s}\blue{w}\purple{e}\orange{r}}}}$

$\large\sf\underline\green{Given:-}$

$\sf\red{ABCD\:is\:a\:rectangle\:in\:which\:diagonal}$

$\sf\purple{AC\:bisects\:∠A\:as\:well\:as\:∠C}$

$\large\sf\underline\pink{Explanation:-}$

(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)

$\sf\green{AB\:=\:BC\:=\:CD\:=\:AD}$

$\sf\blue{Thus,\:ABCD\:is\:a\:square}$

(ii) In ΔBCD,BC = CD

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

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

⇒ ∠CBD = ∠ABD

$\sf\pink{Thus,\:BD\:bisects\:∠B}$

$\large\sf\underline\red{Now,}$

∠CBD = ∠ADB

⇒ ∠CDB = ∠ADB

$\sf\orange{Thus,\:BD\:bisects\:∠D}$