Let $ABC$ be a triangle. Let $T$ be its circumcircle and let $I$ be its incenter. Let the internal bisectors of $A,B,C$ meet $T$ at $A',B',C'$ respectively. Let $B'C'$ intersect $AA'$ at $P$ and $AC$ at $Q$. Let $BB'$ intersect $AC$ at $R$. Suppose that the quadrilateral $PIRQ$ is a kite, i.e. $PI=IR$ and $QP=QR$: how to prove that triangle $ABC$ is a equilateral triangle?
On
Call $\angle CAA'=a,\angle BCC'=b,\angle CBB'=c$. We'll prove that $a=b=c$. First notice that $$\angle PAR= \angle PB'R.$$ This follows directly from the fact that $IPQR$ is a kite. (For example notice that $PAQ$ and $QRB'$ are congruent, since $PQ=QR$, $\angle QRB'=\angle QPA$ and $\angle PQA=\angle RQB'$.) This implies that $ABC$ is isosceles, since $a=\angle PAR=\angle PB'R=\angle BCC'=b$. To conclude we only have to prove that $c=\angle CBB'=\angle ACC'=b$. Now notice that $$\angle CAB'= \angle CBB'$$ because they are subtended by the same chord $B'C$. The last step is to notice $PQA$ and $QRB'$ are congruent triangles. This implies that $AQ=QB'$ and that $AQB'$ is isosceles. From this we deduce that $\angle AB'C'=\angle CAB'$, but it's easy to see that $ \angle AB'C'=\angle ACC'$, and so $$b=\angle ACC'=\angle AB'C'=\angle CAB'=\angle CBB'=c $$ that was the desired equality.
I know that the notation is not easy to read, I'll try to improve it.
Interesting question. For first, we have that $A',B',C'$ are the midpoints of the arcs $BC,AC,AB$ in $T$, hence $AA'\perp B'C'$ and $AP=PI$. Moreover, $$ IP = AP = AC'\sin\widehat{AC'P} = 2R\sin\frac{C}{2}\sin\frac{B}{2}. \tag{1}$$ By applying the sine theorem to the triangle $AIR$, then the bisector theorem, we get: $$ \frac{IR}{\sin\frac{A}{2}} = \frac{AR}{\sin\widehat{AIR}} = \frac{\frac{bc}{a+c}}{\sin\frac{\pi-C}{2}}.\tag{2}$$ We also have: $$ PQ = AP\tan\frac{A}{2},\qquad AQ=\frac{AP}{\cos\frac{A}{2}}\tag{3}$$ as well as: $$ QR = AR-AQ = \frac{bc}{a+c}-\frac{AP}{\cos\frac{A}{2}}. \tag{4}$$
Given $(1),(2),(3),(4)$, to check that $IP=IR$ together with $QP=QR$ imply $a=b=c$ is just horribly tedious but not that difficult.
An alternative, faster solution. If $PQRI$ is a kite, then $\widehat{QB'R}=\widehat{PAQ}$, hence $\frac{\widehat{C}}{2}=\frac{\widehat{A}}{2}$ and $AB=BC$. On the other hand, given that $AA'\perp B'C'$, if $PQRI$ is a kite then $BR\perp AC$, so the angle bisector from $B$ is also the height relative to the $AC$ side, and $B,I,O$ are collinear. Let $\theta=\frac{\widehat{A}}{2}=\frac{\widehat{C}}{2}$: $IR=IP$ implies $\frac{b}{2}\tan\theta = AC'\sin(\theta)$, or $R\sin(\widehat{B})\tan\theta = 2R \sin^2(\theta)$, from which $\sin(4\theta)= \sin(2\theta)$ and $\theta=\frac{\pi}{6}$.