Question

Hello Guys !!

Proof that The Angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of circle !!

Answer 1

Figure in the attachment : Please look at the beautiful attachment : [ I copied the image I am bad at drawing sorry ]

If we join CO :

In Δ COA,

OC = OA [ radius ]

== > ∠CAO = ∠OCA [ base ∠s of an isosceles Δ ]

== > ∠CAO + ∠OCA + ∠COA = 180° [ ∠ sum property of Δ ]

== > ∠OCA + ∠OCA + ∠COA = 180° [ ∠OCA = ∠CAO ]

== > 2 ∠OCA + ∠COA = 180° .....................................(1)

Similarly in Δ COB ,

== > 2 ∠OCB + ∠COB = 180° [ using the above process ]

.........................................(2)

Adding (1) and (2) we get :

== > 2 ∠OCA + 2 ∠OCB + ∠COA + ∠COB = 180° + 180°

[ 2 ( ∠OCA + ∠OCB ) = 2 ∠ACB ................. ]

== > 2 ∠ACB + ∠COA + ∠COB = 360° [ using the above ]

[ ∠COA + ∠COB = reflex ∠AOB = 360° - ∠ AOB ]

== > 2 ∠ACB + 360° - ∠AOB = 360°

== > 2 ∠ACB - ∠AOB = 0 [ Cancelling 360° both sides ]

== > 2 ∠ACB = AOB [ Proved ]

Hence angles subtended at the centre is double the angle at any part on the circles .

Hope it helps :)

#JISHNU#

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Answer 2

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Given : PQ is a chord where (i) PQ is minor arc, (ii) PQ is semicircular arc and (iii) PQ is major arc.

To prove : < POQ = 2 < PRQ.

To construct : Extend OR passing through center of circle.

Proof : Please refer to attachment above.

Concept used : Exterior angle property whivh states that sum of two interior opposite angle is equal to one exterior angle.

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