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Question 173
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
Solution
Angle BQP = Angle BAP
Angle BQP = Half of angle A
Similarly, Angle BQR = half of angle C
Angle BQP + Angle BQR = 0.5 * (A + C) = 0.5 * (180 - B) = 90 - B/2.
Option A is the right answer.
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