Question

Prove that angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Explain with complete calculations & justifications. [Include figure in answer]

Points : 25 ☺

Answer 1

hope it helps you
if you like it then Mark my answer as brainlieast

Answer 2

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)
(proved)