Question

Prove that angle substended by an arc is double the angle substended by it at any point on the remaining part of the circle​

Answer 1

Answer:

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Answer 2

★ Given:-

• A circle with center O.

• PQ substends POQ at the centre.

• PRQ at any point Q on the remaining part of the circle.

★ To Prove:-

• ∠ POQ = 2 ∠PRQ

★ Construction:-

• Join PQ

• Join RO

• Produce RO to a point S on the circle

★ Proof:-

As we know,

When one side of a triangle is produced then the exterior angle so formed is equal to the sum of the interior opposite angles.

➥ So,

In ∆ PRO

∠ 1 + ∠3 = ∠5 ------ eq.1

As,

➥ OR and OP radius of circle

OR = OP

➟ ∠1 = ∠3 ------ eq. 2

➥ Putting eq. 2 in eq. 1

➟ ∠1 + ∠1 = 5

➟ 2∠1 = ∠5 -------- eq.3

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Similarly,

➥ In ∆ ROQ

∠2 + ∠4 = ∠6 ------- eq. 4

➥ As,

As,OR and OQ radius of circle

OR = OQ

➟ ∠2 = ∠4 ------- eq.5

➥ Putting eq. 5 in eq. 4

∠2 + ∠2 = ∠6

➟ 2∠2 = 6 ------ eq.6

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➥ Adding eq. 3 and eq. 6

2∠1 + 2∠2 = ∠5 + ∠6

2( ∠1 + ∠2 ) = ∠5 + ∠6

2PRQ = POQ

Hence Proved.

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