Let $WI$ meet $ABC$ at $T$. Let $D$ be midpoint of arc $BC$. Note that $\angle ITD = 90 = \angle DIE$ so $DIT$ and $BIC$ are tangent at $I$ so by radical axis on $DIT,BIC,ABC$ we have that $EF,BC,DT$ meet at $K$. Since $HW$ is the symmedian of $BHC$ so $BC$ is symmedian of $CHI$ so $K$ is tangent to $BIC$ and since $KH^2 = KB.KC = KE.KF$ we have that $KH$ is tangent to $EFH$. Let $B',C'$ be midpoint of arcs $AC,AB$. Let $HI$ meet $HEF$ at $P$. Note that $B'C'$ is perpendicular bisector of $AI$. Claim $: B',P,C'$ are collinear. Proof $:$ Note that $PI.IH = IE.IF = IC'.IC$ so $PCHC'$ is cyclic so $\angle C'PH = \angle ICH = \angle KIH$ so $C'P \parallel EF \parallel C'B'$. Now since $\angle C'PH = \angle ICH = \angle PHK$ we have that $C'P$ is tangent to $EFH$ so $B'C'$ is tangent to $EFH$.

??! consider the unique circle passing thru E,F and the midpoint of IW. we claim the perpendicular bisector of AI is tangent to this circle. the perpendicular bisector of AI passes thru the midpoint of IW since IAW is right, then it suffices to prove that the tangent to the circle at the midpoint of IW is perpendicular to AI. of course, this is equivalent to proving that the midpoint of IW is symmetric around perpendicular bisector EF, meaning it is symmetric around perpendicular bisector AW since AW and EF are parallel chords of the same circle. of course, this is obvious since it is the circumcenter of IAW. we now prove EFH contains the midpoint of IW, finishing. since WHJ and JAW are both right, we have AHJW cyclic. let the midpoint of IW stop being an illegal and receive the name X. then we have IA * IJ = IH * IW, so let M be midpoint of arc BC and we get IA * IM = IH * IX, now A, M both lie on ABC so we have IE*IF = IA * IM = IH * IX, done.

nAalniaOMliO wrote: In a triangle $ABC$ point $I$ is the incenter, $I_A$ - excenter, $W$ - midpoint of the arc $BAC$ of circumcircle $\omega$ of $ABC$. Point $H$ is the projection of $I_A$ on $IW$. The tangent line to the circumcircle $BIC$ in point $I$ intersects $\omega$ in $E, F$. Prove that the perpendicular bisector to $AI$ is tangent to the circumcircle $EFH$ $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ Let $M$ be the midpoint of the arc $BC$ that does not contain $A$ Let $T$ be the point of tangency of the $A-$mixtilinear incircle with $\omega$ Let $Q\equiv EF\cap BC, P\equiv HI\cap (EFH)$ It is well known that $T\in WI$ Since $\angle MTI=90=\angle IHI_a\Rightarrow HI_a//TM\Rightarrow \frac{IM}{MI_a}=\frac{IT}{MH}$, It is well known that $IM=MI_a\Rightarrow IT=TH...(I)$ $\Rightarrow MT$ is the perpendicular bisector of $IH\Rightarrow MI=MH=MB=MC\Rightarrow BICH$ is cyclic Since $W, I, H$ collinear $\Rightarrow$ the polars of $W, I$ and $H$ wrt $(BICH)$ concur, It is well known that: the polar of $W$ wrt $(BICH)$ is $BC$, the polar of $I$ wrt $(BICH)$ is $EF \Rightarrow$ The tangent through $H$ to $(BICH)$ passes through $Q..(II)$ $EF$ is the radical axis of $(EFH)$ and $\omega$, $BC$ is the radical axis of $(BICH)$ and $\omega \Rightarrow$ The radical axis of $(EFH)$ and $(BICH)$ passes through $H$ and $Q$, By $(II)$ is the tangent through $H$ to $BICH\Rightarrow (EFH)$ and $(BICH)$ are tangent$...(III)$ Since $EF$ is tangent to $(BICH)$ and $(III) \Rightarrow HI$ is the angle bisector of $\angle EHF\Rightarrow P$ is the midpoint of the arc $EF$ that does not contain $H$, Since $I\in EF$(Radical axis of $\omega$ and $(EFH)$) $\Rightarrow IT.IW=IP.IH$, by $(I)\Rightarrow IT.IW=2IT.IH\Rightarrow P$ is the midpoint of $IW$ $\Rightarrow $The tangent through $P$ to $(EFH)$ is parallel to $EF$ and passes through the midpoint of $AI\Rightarrow $The perpendicular bisector of $AI$ is tangent to $(EFH)_\blacksquare$ $\color{blue}\rule{24cm}{0.3pt}$