<p>Let ABC be a triangle with circumcircle Γ and incenter I. Let the internal angle bisectors of ∠A,∠B,∠C meet Γ in A′,B′,C′ respectively. Let B′C′ intersect AA′ at P, and AC in Q. 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. Please give solution with figure.</p>
<p>I R Q P is a kite </p><p>=> angle PQA = angle QRB (vertically opposite)</p><p> angle QPI = angle QRI (kite) => angle APQ = angle QRB</p><p>=>triangle APQ = triangle B'RQ (ASA)</p><p>=> angle A/2 =angle C/2 => angle A = angle C</p><p>ΔABC is isosceles, i.e.</p><p>AB = BC and also ΔIAC is isosceles</p><p>so IA = IC and</p><p>ΔA I Q ≅ ΔB' I Q => AI = B'I = CI i.e. I circumcentre as well incentre ΔABC is equilateral</p>
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