ANSWER:

(i) GIVEN : ABCD is rectangle

To prove : Each angle of rectangle =90

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Proof:

In a rectangle opposite angles of a rectangle are equal

So, ∠A=∠C and ∠B=∠C

But, ∠A+∠B+∠C+∠D=360

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[Sum of angles of a quadrilateral]

∠A+∠B+∠A+∠B=360

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2(∠A+∠B)=360

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(∠A+∠B)=360

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/2

∠A+∠B=180

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But, ∠A=∠B

So, ∠A=∠B=90

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Thus,

∠A=∠B=∠C=∠D=90

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Hence, each angle of a rectangle is 90

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.

(ii) Given : In quadrilateral ABCD, we have

∠A=∠B=∠C=∠D

To prove: ABCD is a rectangle

Proof:

∠A=∠B=∠C=∠D

But these are opposite angle of the quadrilateral.

So, ABCD is parallelogram

And, as angleA=∠B=∠C=∠D

Therefore, ABCD is a rectangle.

(iii) Given : ABCD is a rhombus in which AC=BD

To prove : ABCD is a square

Proof:

JOIN, AC and BD.

Now, in ΔABC and ΔDCB we have

∠AB=DC [SIDES of a rhombus]

∠BC=∠BC [Common]

∠AC=∠BD [Given]

So, ΔABC≅ΔDCB by S.S.S. axiom of of congruency

Thus,

∠ABC=∠DBC [By C.P.P.T.]

But these are made by transversal BC on the same side of parallel lines

AB and DC

So, angleABC+∠DBC=180

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∠ABC=90

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Hence, ABCD is a square.(iv) Given : ABCD is rhombus

To prove : Diagonals AC and BD bisects ∠A,∠C,∠B and ∠D respectively

Proof:

In ΔAOD and ΔCOD, we have

AD=CD [sides of a rhombus are all equal]

OD=OD [Common]

AO=OC [Diagonal of rhombus bisect each other]

So, ΔAOD≅ΔCOD by S.S.S. axiom of congruency

Thus,

∠AOD=∠COD [By C.P.P.T.]

So, ∠AOD+∠COD=180

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[Linear pair]

∠AOD=180

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∠AOD=90

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And, ∠COD=90

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Thus,

OD⊥AC⟹BD⊥AC

Also, ∠ADO=∠CDO [By C.P.C.T.]

So,

OD bisect ∠DDB bisect ∠D

Similarly, we can prove that BD bisect ∠B and AC bisect the ∠A and ∠C.

Step-by-step explanation:

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