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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.

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1 Answer 1

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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.

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  • $\begingroup$ Hi, I can see that $\angle BXA=\angle BIA=90^\circ+\angle C/2$. But how does that give us $\angle PXY=90^\circ-\angle C/2$? $\endgroup$
    – aarbee
    Commented May 23, 2023 at 5:21
  • $\begingroup$ Because $\angle PXY=180^\circ-\angle BXA=180^\circ-(90^\circ+\angle C/2)=90^\circ-\angle C/2$. $\endgroup$
    – oVlad
    Commented May 23, 2023 at 16:17
  • $\begingroup$ 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? $\endgroup$
    – aarbee
    Commented May 24, 2023 at 7:07
  • $\begingroup$ Yes, I meant to write $\angle PIY$ thank you $\endgroup$
    – oVlad
    Commented May 24, 2023 at 7:18
  • $\begingroup$ I understood that X, P, I, Q, Y lie on the same circle. But not able to understand $\angle IPQ=\angle IYQ=\angle ICA=\angle C/2$? $\endgroup$
    – aarbee
    Commented May 24, 2023 at 7:28

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