In the figure, the bisectors of DABC and DACB intersect each other at O. 
Prove that: ∠BOC = 90° + ∠A

In DABC, Since BO is the bisector of ∠ABC
       .^.. ∠1 = ∠2
Similarly ∠3 = ∠4
and ∠A + ∠B + ∠C = 180° ….(i)
Now in DBOC, we have
∠2 + ∠BOC + ∠3 = 180° ….(ii)
Multiplying equation (ii) by 2, we get
2 ∠2 + 2∠BOC + 2∠3 = 2 x 180 = 360°
⇒ ∠B + 2∠BOC + ∠C = 360°
               ⇒ 2∠BOC = 360° - (∠B +∠C)
               ⇒ 2∠BOC = 360° - (180° - ∠A)
                             = 360° - 180° +∠A
               ⇒ 2∠BOC = 180° + ∠A
                  ⇒ ∠BOC = 90° + ∠A