The sides `A B` and `A C` of a `A B C` are produced to `P` and `Q` respectively. If the bisectors of `/_P B C` AND `/_Q C B` intersect at `O ,` then `/_BOC=90^0-1/2/_A` GIVEN : A ` A B C` in which sides `A B` and `A C` are produced to `P` and `Q` respectively. The bisectors of `/_P B C` and `/_Q C B` intersect at `Odot`

Explanation

Given: Bisectors of angle `PBC` and angle `QCB` intersect at `O`.
To prove: `∠BOC=90^@- 1/2∠A`
Proof: `∠BAC=∠A`
Exterior angle`=`Sum of opposite interior angles
`∠PBC=∠C+∠A`
`∠QCB=∠B+∠A`
`∠CBO=(1/2)∠PBC=(1/2)(∠C+∠A)`
`∠BCO=(1/2)∠QCB=(1/2)(∠B+∠A)`
`∠CBO+∠BCO+∠BOC=180^@` (Sum of angles of a triangle)
`(1/2)(∠C+∠A)+(1/2)(∠B+∠A)+∠BOC=180^@`
`(1/2)(∠C+∠A +∠B)+(1/2)∠A+∠BOC=180^@`
`(1/2)180^@+1/2∠A+∠BOC=180^@`
`90^@+1/2∠A+∠BOC=180^@`
`∠BOC=90^2-1/2∠A`
Hence Proved