Given: Chord AB of a circle with centre
O, is produced to C such that BC = OB. CO is joined and produced to meet the circle in D. ∠ACD = y and ∠AOD = x.
To Prove: x = 3y
Proof: In ∆BOC
∵ BC = OB
∴ ∠BOC = ∠BCO
| Angles opposite to equal sides of a triangle are equal
⇒ ∠BOC = y ...(1)
In ∆BOC,
∠OBA = ∠BOC + ∠BCO
| ∵ An exterior angle of a triangle is equal to the sum of its two interior opposite angles
= y + y
= 2y ...(2)
In ∆OAB,
∵ OA = OB
| Radii of the same circle
⇒ ∠OAB = ∠OBA
| Angles opposite to equal sides of a triangle are equal
⇒ ∠OAB = 2y ...(3)
Now, ∵ DOC is a straight line
∴ ∠AOD + ∠AOB + ∠BOC = 180°
⇒ x + {180° - (∠OAB + ∠OBA)} + y = 180°
| Angle sum property of a triangle
⇒ x + 180° - (2y + 2y) + y = 180°
⇒ x = 3y