TITLE: Question on circle and equilateral triangles
QUESTION [5 upvotes]: 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?

REPLY [3 votes]: 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.