Let $\triangle ABC$ be a triangle. Let $Γ$ be its circumcircle, and let $I$ be it’s incenter. 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 $\triangle ABC$ is an equilateral triangle.

I can prove upto isosceles triangle. How to prove equilateral?

  • $\begingroup$ If you can show the base angles of the isosceles triangle are $60$ degrees, you're done. $\endgroup$ – mattos Dec 13 '15 at 13:50

enter image description here

Hope you can complete your task with the aid of the above diagram.

Note: (1) $O$ will lie on the line $BIRB'$ making $BB'$ the diameter of the red circle; and (2) $QB' = QA$ provides further assistance in making each green marked angle (irrespective of their degree of shades) $= 30^0$.


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