In the following figure, sides AB and AC of ΔABC are produced to P and Q respectively. The bisectors of ∠PBC and ∠QCB intersect at O. Then ∠BOC is equal to?
A
1/2 (∠BAC + ∠OCQ)
B
90° + 1/2 ∠BOC
C
90° – 1/2 ∠BAC
D
1/2 (∠PBC + ∠QCB)
Correct Option:
C
Explanation
The external angle bisectors at B and C meet at the excentre opposite A. A standard triangle result gives the angle between these two external bisectors as
90° - A/2.
Since A = ∠BAC,
∠BOC = 90° - 1/2∠BAC.
This is the external-bisector counterpart of the familiar incenter relation 90° + A/2.
