Let $ABC$ be a triangle with an obtuse angle $A$ and incentre $I$. Circles $ABI$ and $ACI$ intersect $BC$ again at $X$ and $Y$ respectively. The lines $AX$ and $BI$ meet at $P$, and the lines $AY$ and $CI$ meet at $Q$. Prove that $BCQP$ is cyclic.
I drew the diagram and saw that angle $AIB$ is equal to angle $AXB$
And angle $AIC$ is equal to angle $AYC$.
I tried making such pairs for the circle $BCQP$ but couldn't do so.
I tried using angle bisector and making some triangles congruent or similar but couldn't conclude.
Best Answer
You can see my diagram here.
Looking in $(ABXI)$ we have $\angle BXA=\angle BIA=90^\circ+\angle C/2$. Thus, $\angle PXY=90^\circ-\angle C/2$.
Further, note that $\angle PIC=90^\circ+\angle A/2$ and looking in $(ACYI)$ we have \begin{align*}\angle YIQ&=\angle YAC=180^\circ-\angle C-\angle AYC= 180^\circ-\angle C-\angle AIC \\ &=180^\circ-\angle C-\angle(90^\circ+\angle B/2)=\angle A+\angle B/2-90^\circ.\end{align*}Therefore, $\angle PIY=\angle PIC-\angle YIQ=90+\angle C/2$ so $\angle PIY+\angle PXY=180^\circ$.
Thus,the points $X,Y,P,I$ lie on a circle and analogously, we may infer that $Q$ is on this circle too. Hence, $\angle IPQ=\angle IYQ=\angle ICA=\angle C/2$ but $\angle ICB=\angle C/2$ as well, hence $P,Q,B,C$ lie on the same circle.