Let D be an arbitrary point inside $\Delta ABC$. Let $\Gamma$ be the circumcircle of $\Delta BCD$. The external angle bisector of $\angle ABC$ meets $\Gamma$ again at $E$. The external angle bisector of $\angle ACB$ meets $\Gamma$ again at $F$. The line $EF$ meets the extension of $AB$ and $AC$ at $P$ and $Q$ respectively. Prove that the circumcircles of $\Delta BFP$ and $\Delta CEQ$ always pass through the same fixed point regardless of the position of $D$. (Assume all the labelled points are distinct.)
Problem
Source: 2020HKTST1 Q2
Tags: geometry, circumcircle, angle bisector
MarkBcc168
24.08.2019 16:26
The fixed point is the $A$-excenter $I_A$. To see that, notice $$\angle I_AFP = \angle I_AFE = \angle EBC = \angle I_ABC =\angle I_ABP$$
ubermensch
24.10.2019 17:18
The external angle bisectors were a total giveaway Anyhow, notice that if $\angle ABC = \beta => \angle PBI_A=90 -\frac{\beta}2$ and $\angle CFE = \angle CBE = 90 - \frac{\beta}2$, thus $BI_AFP$ (and similarly $CQEI_A$) is a cyclic quadrilateral, thus our fixed point is $I_A$.
Mogmog8
02.06.2022 05:22
We claim $I_A,$ the $A$-excenter of $\triangle ABC,$ lies on both $(BFP)$ and $(CEQ).$ Indeed, $$\measuredangle PBI_A=\measuredangle I_ABC=\measuredangle EFC=\measuredangle PFI_A$$and $I_A\in(CEQ)$ follows similarly. $\square$
jasperE3
08.06.2022 17:14
Let $I_A$ be the $A$-excenter and use directed angles, then: $$\measuredangle PFI_A=\measuredangle EFC=\measuredangle EBC=\measuredangle PBI_A$$so $I_A$ lies on $\odot(BFP)$. Similarly, $I_A$ lies on $\odot(CEQ)$.