Question

Prore that the angle Subtended by an are at the contre is double the angle subtended by it
angle subtended by it at
any point ?​

Answer 1

Answer:

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)