In figure, the bisectors of the exterior angles at B and C formed by producing sides AB and AC of Δ ABC intersect each other at the point O. Prove that ∠ BOC=90° - ∠ A .

Given: ABC is a triangle . Let AB be produced to P and AC be produced to Q. Let the bisectors of ∠ PBC and  ∠ QCB meet at O.
To prove: ∠ BOC = 90° -
Proof: In Δ ABC,
∠ A + ∠ B + ∠ C = 180° (Sum of the three angles of a triangle is 180° )
∠ A + ∠ B + ∠ C = × 180°
∴ = 90° - ……………………….(1)
= 90° -
In Δ BOC,
∠ BOC + ∠ CBO + ∠ BCO = 180°
∠ BOC = 180° - (∠ CBO + ∠ BCO) ………………(2)
∠ PBC = 180° - ∠ ABC (Since ∠ ABC and ∠ CBP form a linear pair)
Dividing by 2,
∠ PBC = (180° - ∠ ABC)
∴ ∠ CBO = 90° …………………..(3)
 Ð QCB = 180° - ∠ ACB (Since ∠ QCB and ∠ ACB form a linear pair)
∠ QCB = (180° -  Ð ACB)
∴ ∠ BCO = 90° - ………………………(4)
Substituting (3) and (4) in (2),
∠ BOC = 180° - (∠ CBO + ∠ BCO)
          = 180° - (90° - + 90° - )
          = 180° - (180° - - )
          =+…………………………….(5)
But from (1), + = 90° -
Substituting in (5),
∠ BOC = 90° -