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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 :
∠ACB = 2∠APR,
Solution
Join PQ and PR
CR is the bisector of ∠ACB
`=> ∠ACR = 1/2 ∠ACB`
Also, ∠ACR = ∠APR
(Angle in the same segment)
∴ ∠ACB = 2∠APR
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