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Let ABC be a triangle with circumcircle Γ and incentre I. Let the internal angle bisectors of ∠A, ∠B, and ∠C meet Γ in A', B' and C' respectively. Let B'C' intersect AA' in P and AC in Q, and let BB' intersect AC in R. Suppose the quadrilateral PIRQ is a kite; that is IP = IR and QP = QR. Prove that ABC is an equilateral triangle.

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I R Q P is a kite

=> ΔABC is isosceles, i.e. AB = BC and also ΔIAC is isosceles

so IA = IC and ΔA I Q ≅ ΔB' I Q

=> AI = B'I = CI

i.e. I circumcentre as well incentre ΔABC is equilateral

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