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

Given: OB is the bisector of ∠ ABC and OC is the bisector of ∠ ACB of Δ ABC.
To prove: 90° + ∠ A
Proof: In a triangle the sum of three angles is 180°  ----------- (1)
In Δ ABC,  
.^.. ∠ A + ∠ B + ∠ C = 180°
Dividing by 2,
 ∠ A + ∠ B + ∠ C = = 90°
[∠ B + ∠ C ] = 90° -  ∠ A    --------------  (2)
In Δ BOC,  
∠ BOC + ∠ OBC + ∠ OCB = 180°   (same as (1)
                          ∠ BOC = 180° - ∠ OBC - ∠ OCB                                  
                        .^..∠ BOC = 180^o - (∠ B+∠ B) ………..……….. (3)
(Since OB, OC are bisector of ∠ ABC, ∠ ACB).
Substitute (2) in (3)
                      .^.. ∠ BOC = 180° - [90° - ∠ A ]
                                    = 180° - 90° + ∠ A
                                    = 90° + ∠ A