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`

Answer 1

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`