(a)   AB is the chord of a circle with centre O and AB is produced to C such that BO = BC.  CO produced meets the circle at D.

If angle ACD = y⁰ and angle AOD = x⁰ , prove that x = 3y. 

Dear Student!

Here is the answer to your query.

Given : AB is a of a chord circle and OB = BC, CO produced meats the circle at D also ∠ACD = y° and ∠AOD = x°

In ΔOBC

OB = BC

∴ ∠BOC = ∠BCO = y°  (Angle opposite to equal sides are equal)  .....(1)

and ∠OBA = ∠BOC + ∠BCO  (exterior angle property)

= y° + y° = 2y°  .......(2)

Now in ΔAOB

AO = OB = radius

∴ ∠AOB = ∠OBA = 2y°

and ∠AOB + ∠OAB + ∠OBA = 180°  (Angle sum property)

⇒ ∠AOB + 2y° + 2y° = 180°  (from (2))

⇒ ∠AOB  = 180° – 4y°  (from (3))

Since COD is a straight line

∴ ∠COD = ∠DOA + ∠AOB + ∠BOC = 180°

⇒ x° + 180° – 4y° + y° = 180°  (from (1) and (3))

⇒ x° – 3y°  = 180° – 180° = 0

⇒ x = 3y

Cheers!