in the figure, chord AB of circle with centre O, is produced to C such that BC = OB . CO is joined and produced to meet the circle at D.If angle ACD = y and angle AOD = x,show that x = 3y.

O is the centre of the circle. AB is the chord of the circle. AB is produced to C such that OB = BC. CO produced intersects the circle in D.

∠OCB = y and ∠AOD = x.

In ΔOBC,

OB = BC (Given)

∴ ∠OCB = ∠BOC  (Equal sides have equal angles opposite to them)

⇒ ∠BOC = y

∠OBA = ∠BOC + ∠OCB  (Exterior angle of a triangle is equal to sum of its interior opposite angles)

∴ ∠OBA = y + y = 2y  ...(1) 

In ΔAOB,

OA = OB (Radius of the circle)

∴ ∠OBA = ∠OAB  (Equal sides have equal angles opposite to them)

⇒∠OAB = 2y [Using (1)]

In ΔAOC,

∠AOD = ∠OAC + ∠OCA  (Exterior angle of a triangle is equal to sum of its interior opposite angles)

∴ x = 2y + y   [∠OAC = ∠OAB]

⇒ x = 3y