In $$\triangle$$​ABC, O is the point of intersection of the bisectors of angle B and angle A. If the angle BOC = 108°, then angle BAO is: 

Correct option is D

Given:

O is the point of intersection of the bisectors of ∠B and ∠A
∠BOC = 108°
Formula Used:
O is the incentre of the triangle.
Angle of the incentre of the triangle is equal to the sum of 90°  and half of the angle opposite to the incentre of a triangle.
Sum of all angles of triangle =180°
Solution:
The point at which angle bisectors meet is the incentre of the triangle.
Hence, O is the incentre of the triangle.
If ∠BOC = 108°

​$$∠BOC = 90° +\frac{ (∠BAC)}{2} \\\ \\108°= 90° +\frac{ (∠BAC)}{2}\\\ \\∠BAC= 36° \\\ \\∠BAO=\frac{ (∠BAC)}{2 }= 18°$$​​