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* = 2*y ...*(1)

In ΔAOB,

OA = OB (Radius of the circle)

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

⇒∠OAB = 2*y* [Using (1)]

In ΔAOC,

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

∴ *x = *2*y + y* [∠OAC = ∠OAB]

⇒ *x* = 3*y *

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