(98 POINTS FOR BRAINLIEST ANSWER) Prove that the angle subtended by an arc at the centre of a circle is double the angle subtended by it at any point. [Should do all 3 cases/proofs.]
Answers
To prove : <AOB = 2 <ACB
Construction : Join CO and produce it to D
(i) OA = OC (Radii)
(ii) <OCA = <OAC
(angles opposite to equal sides are equal.)
(iii) In ΔAOC
<AOD = <OCA + <OAC
(Exterior angles of a triangle = Sum of the interior opposite angles)
(iv) <AOD = <OCA + <OCA
(substituting <OAC by <OCA)
(v) <AOD = 2 <OCA (by addition)
(vi) Similarly in triangle BOC
<BOD = 2 <OCB
(vii) <AOD + <BOD = 2 <OCA + 2<OCB
= 2(<OCA + <OCB)
(<AOD + <BOD = <AOB and <OCA + <OCB = <ACB)
(viii) <AOB = 2 <ACB
Answer:
Step-by-step explanation:
Report by GovindKrishnan 12.03.2017
Answers
THE BRAINLIEST ANSWER!
maria9
Maria9 Ace
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)