In the figure, the sides AB and AC of a triangle ABC are produced to P and Q respectively. The bisectors of ∠PBC and ∠QCB intersect at O. Prove that ∠BOC = 90° - ∠A.

Since BO and CO are angle bisectors of ∠PBC and ∠QCB respectively 
.^.. ∠1 = ∠2 and ∠3 = ∠4
Side AB and AC of DABC are produced to P and Q respectively .
Hence Exterior ∠PBC = ∠A + ∠C ….(i)
  and Exterior ∠QCB = ∠A + ∠B ….(ii)
Adding equations (i) and (ii), we get
∠PBC + ∠QCB = 2∠A + ∠B + ∠C
=> 2∠2 + 2∠3 = ∠A +(∠A + ∠B + ∠C)
=> 2∠2 + 2∠3 = ∠A + 180°  
   ⇒ ∠2 + ∠3 = 90° + ∠A ….(iii)
Now in ΔBOC, ∠2 + ∠BOC + ∠3 = 180° ….(iv) 
From equations (iii) and (iv), we get
∠BOC + 90° + ∠A = 180°
\\ ∠BOC = 90° - ∠A