In the given figure, AB = AC. If BO and CO, the bisectors of ∠B and ∠C respectively meet at O and BC is produced to D, prove that ∠BOC = ∠ACD.

In △ABC,

AB = AC

⇒ ∠B = ∠C = x (let) (Angles opposite to equal sides in a triangle are equal)

From figure,

∠B = ∠ABO + ∠OBC

⇒ x = ∠OBC + ∠OBC (∵ ∠ABO = ∠OBC, as OB is the bisector of angle B)

⇒ x = 2∠OBC

⇒ ∠OBC = $\dfrac{\text{x}}{2}$ ….(1)

From figure,

∠C = ∠ACO + ∠OCB

⇒ x = ∠OCB + ∠OCB (∵ ∠ACO = ∠OCB, as OC is bisector of angle C)

⇒ x = 2∠OCB

⇒ ∠OCB = $\dfrac{\text{x}}{2}$ ….(2)

From eq.(1) and (2), we have :

⇒ ∠OCB = ∠OBC

In △BOC,

By angle sum property of triangle,

⇒ ∠OCB + ∠OBC + ∠BOC = 180°

⇒ $\dfrac{\text{x}}{2} + \dfrac{\text{x}}{2}$ + ∠BOC = 180°

⇒ x + ∠BOC = 180°

⇒ ∠BOC = 180° - x ….(3)

From figure,

⇒ ∠ACB + ∠ACD = 180° (Linear pair)

⇒ ∠ACO + ∠OCB + ∠ACD = 180°

⇒ $\dfrac{\text{x}}{2} + \dfrac{\text{x}}{2}$ + ∠ACD = 180° (∵ ∠ACO = ∠OCB)

⇒ x + ∠ACD = 180°

⇒ ∠ACD = 180° - x ….(4)

From eq.(3) and (4), we have:

⇒ ∠BOC = ∠ACD

Hence, proved that ∠BOC = ∠ACD.