Considering the △ ABC
It is given that AB = AC
So we get
∠ ABC = ∠ ACB
Dividing by 2 both sides
½ ∠ ABC = ½ ∠ ACB
So we get
∠ OBC = ∠ OCB …… (1)
By using the angle sum property in △ BOC
∠ BOC + ∠ OBC + ∠ OCB = 180o
Substituting equation (1)
∠ BOC + 2 ∠ OBC = 180o
So we get
∠ BOC + ∠ ABC = 180o
From the figure we know that ∠ ABC and ∠ ABP form a linear pair of angles so we get
∠ ABC + ∠ ABP = 180o
∠ ABC = 180o – ∠ ABP
By substituting the value in the above equation we get
∠ BOC + (180o – ∠ ABP) = 180o
On further calculation
∠ BOC + 180o – ∠ ABP = 180o
By subtraction
∠ BOC – ∠ ABP = 180o – 180o
∠ BOC – ∠ ABP = 0
∠ BOC = ∠ ABP
Therefore, it is proved that the exterior angle adjacent to ∠ ABC is equal to ∠ BOC.