prove the angle subtended by an arc at the centre is twice the angle the angle in remaining part of circle
Answers
ACB ar any point C on the remaining part of the circle .
TO PROVE :- /_ AOB = 2( /_ ACB)
CONSTRUCTION :- join CO and produce it to any point D
PROOF :-
OA = OC [radii of same circle ]
/_ OAC = /_ ACO
[angles opp to equal side's of a triangle are equal]
/_ AOD = /_OAC + /_ACO
[ext angles = sum of equal opp angles]
/_AOD = 2(/_ACO)-------------(1)
[/_OAC = /_ACO]
similarly,
/_ DOB = 2(/_OCB) -------------(2)
In fig (i) and (iii)
adding (1) And (2)
/_AOD + /_ DOB = 2(/_ACO) + 2(/_OCB)
/_AOD + /_ DOB = 2(/_ACO + /_OCB)
/_AOB = 2(/_ACB)
In fig (ii)
subtracting (1) from (2)
/_DOB - /_DOA = 2(/_OCB - /_ACO)
/_AOB = 2(/_ACB)
hence in all cases we see
/_AOB = 2(/_ACB)
(proved)
hope u like it
Answer:
Given :
An arc PQ of a circle subtending angles POQ at the centre O and PAQ at a point A on the remaining part of the circle.
To prove : ∠POQ=2∠PAQ
To prove this theorem we consider the arc AB in three different situations, minor arc AB, major arc AB and semi-circle AB.
Construction :
Join the line AO extended to B.
Proof :
∠BOQ=∠OAQ+∠AQO (1)
Also, in △ OAQ,
OA=OQ [Radii of a circle]
Therefore,
∠OAQ=∠OQA [Angles opposite to equal sides are equal]
∠BOQ=2∠OAQ (2)
Similarly, BOP=2∠OAP (3)
Adding 2 & 3, we get,
∠BOP+∠BOQ=2(∠OAP+∠OAQ)
∠POQ=2∠PAQ (4)
For the case 3, where PQ is the major arc, equation 4 is replaced by
Reflex angle, ∠POQ=2∠PAQ
