Question

In the figure, given below, AB = AC.

Prove that: ∠BOC = ∠ACD.

Answer 1

Let  ∠ABO = ∠OBC = x and ∠ACO = ∠OCB = y

In ΔABC,

∠BAC = 180° - 2x - 2y          ...(i)

Since, ∠B = ∠C                 ...[AB = AC]

`1/2"B" = 1/2"C"`

⇒ x = y

Now, 

∠ACD = 2x + ∠BAC         ...[Exterior angle is equal to sum of opp. interior angles]

∠ACD = 2x + 180° - 2x - 2y      ...[From (i)] 

∠ACD = 180° - 2y                      ...(i)

In ΔOBC,

∠BOC = 180° - x - y 

⇒ ∠BOC = 180° - y - y            ...[Already proved]

⇒ ∠BOC = 180° - 2y              ...(ii)

From (i) and (ii)

∠BOC = ∠ACD