ABC is a cyclic triangle and the bisectors of ∠BAC, ∠ABC and ∠BCA meet the circle at P, Q and R respectively. Then the angle ∠RQP is:

A. $${90^ \circ } - \frac{{\text{B}}}{2}$$

B. $${90^ \circ } + \frac{{\text{C}}}{2}$$

C. $${90^ \circ } - \frac{{\text{A}}}{2}$$

D. $${90^ \circ } + \frac{{\text{B}}}{2}$$

Answer: Option A

Solution (By Examveda Team)

From chord BP
∠BAP = ∠BQP = z
(Angle subtended by same chord on circumference)
Similarly from chord RB
∠RCB = ∠RQB = y
In ΔABC
2x + 2y + 2z = 180°
x + y + z = 90°
y + z = 90° - x
∠RQP = 90° - x
∠RQP = $${90^ \circ } - \frac{{\angle {\text{B}}}}{2}$$