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Question
A triangle ABC is inscribed in a circle. The bisectors of angles BAC, ABC and ACB meet the circumcircle of the triangle at points P, Q and R respectively. Prove that :
`∠QPR = 90^circ - 1/2 ∠BAC`
Solution
Join PQ and PR
Adding (i) and (ii)
We get
∠ABC + ∠ACB = 2(∠APR + ∠APQ) = 2∠QPR
`=>` 180° – ∠BAC = 2∠QPR
`=> ∠QPR = 90^circ - 1/2 ∠BAC`
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