Skip to main content

Theorem | the angle subtended by an arc of a circle | at the centre is double the angle subtended by it at | any point on the remaining part of the circle

Theorem :

A circle  with centre O in which AB subtends AOB at centre and angle ACB at any point  on the remaining part of the circle.
R.T.P. - AOB =  2 xACB

Solution :

Construction: Join CO and produce CO to point D .

1. In AOC, OC = OA || Radii of same circle .
2. OCA= OAC || Base angles of isoscles ∆ in which OC = OA
3. Exterior AOD = OCA+ OAC || Exterior ∠ of a ∆ = sum of two opposite interior angles.
or AOD = OCA+ OCA
or AOD = 2 . ∠ OCA - - - - Eqn. 1
4. Similary ot can be proved that BOD = 2. OCB - - - - - - Eqn. 2
5. Adding Eqn 1 and Eqn. 2 we get
AOD + BOD = 2 . ∠OCA + 2. ∠OCB
AOB = 2 (∠OCA + ∠OCB)
AOB = 2 ∠ACB Proved
Video will be loaded soon.

You may also Like These !

Comments



Weekly Popular

Question Answer of 'Daffodils' | English Literature

Question Answers 'The Happy Prince' - English Literature | Reference to Context