# Using angle bisector to prove a quadrilateral is concyclic.

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.

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.

• Hi, I can see that $\angle BXA=\angle BIA=90^\circ+\angle C/2$. But how does that give us $\angle PXY=90^\circ-\angle C/2$? May 23 at 5:21
• Because $\angle PXY=180^\circ-\angle BXA=180^\circ-(90^\circ+\angle C/2)=90^\circ-\angle C/2$. May 23 at 16:17
• You have written "Therefore, $\angle PIQ=\angle PIC-\angle YIQ$" But aren't angle PIQ and PIC just the same? Maybe instead of PIQ, you wanted to write PIY? May 24 at 7:07
• Yes, I meant to write $\angle PIY$ thank you May 24 at 7:18
• I understood that X, P, I, Q, Y lie on the same circle. But not able to understand $\angle IPQ=\angle IYQ=\angle ICA=\angle C/2$? May 24 at 7:28