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
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let there be a circle with center O . arc AB intends AOB at the center and 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)
Hence proved
