Nice problem! The $\sqrt{bc}$ reduction feels a bit underwhelming, though (and it might be how the problem was created?). We $\sqrt{bc}$ invert. Then $H$ maps to $A'$, while $O$ maps to the reflection of $A$ across $BC$, which we label $R$. Additionally, $F$ maps to the intersection between the circle centered at $A$ with radius $AR$, which we label $S$. Finally, the intersection of $\left(AFH\right)$ and $\left(BOC\right)$ maps to $K$, the second intersection of $A'S$ and $\left(BRC\right)$. The conclusion thus reduces to showing that $\angle AHK=90^{\circ}$, or equivalently $KH\parallel BC$. This further reduces to $\angle CSA'=\angle OAC$. Now note that $A$ is the $A'$-excenter of $\triangle A'BC$. Therefore (by phantom points), if rephrased with respect to $\triangle A'BC$, the problem reduces to the following. Rephrasing wrote: Let $ABC$ be a triangle and let $I_A$ be its $A$-excenter. Let $S$ be a point so that $\angle BSA=\frac{1}{2}\angle ACB$ and $\angle CSA=\frac{1}{2}\angle ABC$, on the same side of $BC$ as $A$. Show that $I_AR$ is twice the $A$-exradius. Let $U$ be the reflection of $I_A$ across $AC$ and let $V$ be the reflection of $I_A$ across $AB$. Then we wish to show that $I_A$ is the circumcenter of $\triangle VSU$. Since $I_A$ lies on the perpendicular bisector of $UV$, it suffices to show that $\angle VSU=180^{\circ}-2\angle VI_AU$. Now it easily follows that $\triangle ACI_A\equiv\triangle ACU$, so $\angle AUC=\frac{1}{2}\angle ABC$. Therefore $ASUC$ is cyclic, so $$\angle CSU=\angle CAU=\angle I_AAC=\frac{1}{2}\angle BAC.$$ Similarly $\angle VSB=\frac{1}{2}\angle BAC$, hence \begin{align*} \angle VSU&=\angle VSB+\angle BSC+\angle CSU\\ &=90^{\circ}+\frac{1}{2}\angle BAC\\ &=180^{\circ}-\angle BI_AC\\ &=180^{\circ}-\frac{1}{2}\angle VI_AU, \end{align*}as desired.

(Diagram to follow...) Let $D$ be the reflection of $O$ over $A$. Suppose $\Gamma$ and the circle on diameter $AA'$ intersect at $P\neq A'$. Invert about $(ABC)$. $\Gamma$ swaps with $BC$, so the image of $A'$ lies on $BC$, which implies $P^*$ is the foot of the altitude from $A$ to $BC$. $D^*$ is the midpoint of $AO$, and $F^*$ is the intersection of the perpendicular bisector of $AO$ with $BC$. So $P^*$ and $D^*$ lie on the circle with diameter $AF^*$, i.e. $DAFP$ is cyclic. So it suffices to show that $H$ also lies on this circle. Let the bisector of $\angle{A}$ intersect $(ABC)$ at $M\neq A$, and let $X$, $Y$ be the centres of $(BHC)$ and $\Gamma$ respectively. These circles swap under root-$b$-$c$ inversion, as $O$ is sent to the reflection of $A$ over $BC$, which lies on $(BHC)$. Thus, $AM$ bisects $\angle{XAY}$. Reflecting $AO$ over $AM$ and $AY$ gives lines $AH$ and $AF$ respectively, so we can compute $\angle{HAF}=\angle{XAY}$. So it suffices to show that $\angle{HDF}=\angle{XAY}$. A homothety of factor $\frac 12$ centred at $A$ takes $(BHC)$ to the nine-point circle of $\triangle{ABC}$, so the midpoint of $AX$ is $N_9$. But this is also the midpoint of $OH$, so $AOXH$ is a parallelogram. Hence, $ADHX$ is also a parallelogram, i.e. $DH\parallel AX$. But also $DF\parallel AY$ as they are both perpendicular to $OF$, so $\angle{HDF}=\angle{XAY}$ as desired.

Rename the point $A'$ in the problem to $D$. We will start with a $\sqrt{bc}$ inversion. This gives us the following problem: Triangle $\triangle ABC$ with orthocenter $H$ and circumcenter $O$. Let $AO$ intersect $(BOC)$ at $D\neq O$ and let $A'$ be the reflection of $A$ in $BC$. $F\neq A'$ is on $(A'BHC)$ such that $AF=AA'$. Prove that the line through $H$ parallel to $BC$ and the line $DF$ intersect at the circle $(A'BHC)$. Because of Thales this is equivalent to proving $\angle A'FD=90^\circ$. Now let $AF$ intersect $(A'BHCF)$ again at $X$. From inversion things we get that$AH\cdot AD=AB\cdot AC=AO\cdot AA'$ and that implies $A'D\parallel HO$. Also obviously $A'F\parallel HX$ because of isosceles trapezoid. Now we see $\frac{AX}{AF}=\frac{AH}{AA'}=\frac{AO}{AD}$. This implies that triangles $\triangle HOX$ and $\triangle A'DF$ are similar with parallel sides. Therefore we want to show that $\angle HXO=90^\circ$. Now we have reduced the problem to the following lemma: Triangle $\triangle ABC$ with circumcenter $O$ and orthocenter $H$. $X\neq H$ is on $(BHC)$ such that $AX=AH$. Prove that $\angle HXO=90^\circ$. Let $M$ be the reflection of $O$ in $BC$, we know that $M$ is the center of $(BHC)$. Let $X'$ be the point such that $AMX'O$ is an isosceles trapezoid. We have $AX'=OM=AH$ and $MX'=AO=BO=MB$, so $X'=X$. Now look at the radical axis of circle $(BHC)$ and the circle centered at $A$ through $H$. Obviously $HX$ is that radax so $HX\perp AM$. Also $AM\parallel XO$, so $HX\perp XO$ and we are done.

no complex bashes yet? Invert around $O$. Then $A'$ goes to $A'^\ast=AO\cap BC$. Then $(AA')$ goes to $(AA'^\ast)$ and $\Gamma$ goes to $BC$, so their other intersection is the foot of the perpendicular from $A$ to $BC$, call it $X$. $F$ goes to the intersection of $BC$ and the perpendicular bisector of $AO$, call it $Y$. Let $H$ go to $Z$, we can directly compute it later. Obviously $\angle AXY=90^{\circ}$ so it suffices to prove $\angle YZA=90^{\circ}$. Now let's bash. I did this on paper so I will only provide a summary here. For $y$, our two equations simplify to \[bc\overline{y}+y=b+c\]\[a^2\overline{y}+y=a.\]Obviously \[z=\frac{1}{\overline{a+b+c}}=\frac{abc}{ab+bc+ca}.\]Now after only five lines of computation, we get \[\frac{y-z}{z-a}=\frac{a(b^2+bc+c^2)}{(b+c)(bc-a^2)}\]which we can check to be equal to its negative conjugate.

$\sqrt{bc}$ inversion gives the following problem. Quote: Let $\triangle ABC$ be a triangle with orthocenter $H$ and circumcenter $O$. Line $AO$ meet $\odot(BOC)$ at $A'$. Let $A_1$ be the reflection of $A$ across $BC$. Point $F$ lies on $\odot(BHC)$ such that $AA_1=AF$. Prove that $A'F$, $\odot(BHC)$, and line through $H$ parallel to $BC$ are concurrent. To prove this, let $O_1$ be the center of $\odot(BHC)$. Note that $A'A_1\parallel OH$, so since $AO_1$ bisects $OH$, it follows that $AO_1$ bisects $A_1A'$. Now, if $M$ is the midpoint of $A_1A'$, then $MF=MA_1=MA'$, which implies $\angle A_1FA'=90^\circ$. Thus, if $A'F$ meet $\odot(BHC)$ again at $P$, then $\angle PHA_1=\angle PFA_1=90^\circ$, so $HP\parallel BC$.

When the easiest problem at RMM is the p5 geo..

Let $AHF$ meet $\Gamma$ at $X$. Let the circle with center $A$ and radius $AO$ meet $AO$ at $E \neq O$. we need to prove $\angle AXA' = 90$. Note that $\angle AXA' = \angle AXF+\angle FXA' = \angle AXF+\angle FOA$. Since $\angle EFO=90$ we need to prove $\angle AEF=\angle AXF$ so we need to prove $E$ lies on $AHFX$. Let $O'$ be reflection of $O$ about $BC$ and $N$ be the midpoint of $OH$. Let $T$ be the reflection of $A$ about $OF$. Let $AT$ meet $OO'$ at $K$. First note that $K$ lies on perpendicular bisectors of $BC$ and $FO$ so $K$ is the center of $BOC$. Note that since $EA=AO$ and $HN=NO$ then $\angle EHA=\angle HAN=\angle OO'A$ and $\angle EFA=\angle FEA=\frac{\angle FAO}{2}=\angle TAO=\angle ATO$ so we need to prove $AOO'T$ is cyclic. we have that $KT.KA=KO^2-R^2$ where $R$ is the radius of $ABC$ and we need to prove $KT.KA=KO'.KO$ so we need to prove $KO.(KO-KO')=R^2$ or $KO.OO'=R^2$ which is true since $OCO'$ and $OKC$ are similar. we're done.

Perform $\sqrt{bc}$ inversion. New Problem Statement: $ABC$ is a triangle with circumcenter $O$ and orthocenter $H$. Let $AO\cap (BOC)=R,\ A'$ be the reflection of $A$ over $BC$. Let $W$ be the circumcenter of $(BHC)$ and $F$ be the reflection of $A'$ over $AW$. Let $K\in (BHC)$ and $HK\parallel BC$. Prove that $R,K,F$ are collinear. Work on the complex plane. Let $(ABC)$ be the unit circle. Assume $a=1$. We have $w=b+c,\ a'=b+c-bc$ hence $k=bc+b+c$. \[f=\frac{(b+c-a)(\frac{1}{b}+\frac{1}{c}-\frac{a}{bc})+\frac{a}{b}+\frac{a}{c}-\frac{b+c}{a}}{\frac{1}{b}+\frac{1}{c}-\frac{1}{a}}=\frac{b^2+c^2+2bc+1-b-c-b^2c-bc^2}{b+c-bc}\]\[r=\frac{bc+1}{b+c}, \ k=bc+b+c\]\[\frac{k-r}{k-f}=\frac{b+c+bc-\frac{1+bc}{b+c}}{b+c+bc-\frac{b^2+c^2+2bc+1-b-c-b^2c-bc^2}{b+c-bc}}=\frac{(b+1)(c+1)(b+c-1)(b+c-bc)}{(b+c)(-b^2c^2+b^2c+bc^2+b+c-1)}\]\[\overline{\frac{(b+1)(c+1)(b+c-1)(b+c-bc)}{(b+c)(-b^2c^2+b^2c+bc^2+b+c-1)}}=\frac{(\frac{b+1}{b})(\frac{c+1}{c})(\frac{b+c-bc}{bc})(\frac{b+c-1}{bc})}{(\frac{b+c}{bc})(\frac{-1+b+c+bc^2+b^2c-b^2c^2}{b^2c^2})}=\frac{(b+1)(c+1)(b+c-1)(b+c-bc)}{(b+c)(-b^2c^2+b^2c+bc^2+b+c-1)}\]As desired.$\blacksquare$

Let $O'$ be the reflection of $O$ over $BC$ and let $(AOO')$ and $(ABC)$ intersect at point $D$. Observe that line $BC$ bisects circle $(AOO')$, which is equivalent to $BC$ and $(AOO')$ forming right angles with one another. then, inverting at $O$ with radius $AO$ sends $(AOO')$ to line $AD$ and sends $BC$ to $(BOC)$, inversion preserves angles so these new two curves must still be perpendicular to each other, thus line $AD$ bisects $(BOC)$. Now recall that $AF=AO$, so the perpendicular bisector of $FO$ both passes through $A$ and bisects $(BOC)$ implying that this is the same line as $AD$, thus $D$ lies on the perpendicular bisector so $FD=DO=AO=AF$. Thus $AODF$ is a rhombus. Reflect $H$ across $BC$ to point $H'$ lying on the circumcircle. Clearly $AH'||OO'$, $AO=H'O=HO'$, and $\angle OO'H=\angle H'OO'=\angle OH'A=\angle H'AO$ (using the fact that triangle $AOH'$ is isosceles) thus $AOO'H'$ is a parallelogram. Thus $\vec{OA}=\vec{DF}=\vec{O'H}$. Now define point $E$ to be the reflection of $O$ over $A$. Clearly, $\vec{AE}=\vec{OA}$, so if we shift cyclic quadrilateral $AODO'$ by vector $\vec{OA}$, we get quadrilateral $EAFH$, so $E$ lies on $(AFH)$. Let $(AFH)$ intersect with $(BFC)$ at $X$, we now have that $\angle AXA'=\angle AXF+\angle FXA'=\angle AEF+\angle FOA'=\angle OEF+\angle FEO+\angle OFE=\angle OFE$ which is clearly a right angle, thus the circle with diameter $AA'$ indeed passes through $X$ as desired.

same as jatloe Let $\Omega = (ABC),$ and let $X,$ $Y,$ and $D$ be on $BC$ such that $AX = OX$,$Y$ lies on $AO,$ and $D$ lies on $AH.$ Invert about $\Omega$; let the point $H$ get sent to $H'.$ Then $\Gamma$ is sent to $BC$ and the circle with diameter $AA'$ is sent to the circle with diameter $AY.$ Furthermore, the circle with center $A$ passing through $O$ is sent to the perpendicular bisector of $AO,$ so $F$ is sent to $X.$ We would like to show that the circle with diameter $AY$ intersects $(AH'X)$ on $BC,$ or equivalently $AH'XD$ is cyclic. Since $\angle ADX = 90^\circ,$ it suffices to show that $\angle AH'X = 90^\circ$ as well. We now use complex numbers with $\Omega$ as the unit circle. Then $h = a+b+c,$ so $$h' = \frac{1}{\overline{a+b+c}} = \frac{abc}{ab+bc+ca}.$$Now we find $x.$ Let $\omega = \frac{1}{2} + \frac{\sqrt{3}}{2} i$, so that the perpendicular bisector of $AO$ passes through $a\omega$ and $a\overline{\omega}.$ Thus by the complex intersection formula, we get $$x = \frac{a \omega \cdot a\overline{\omega} (b+c) - bc(a\omega + a\overline{\omega})}{a\omega \cdot a\overline{\omega} - bc} = \frac{a^2 b + a^2 c - abc}{a^2 - bc}.$$Then we would like to show that $$z = \frac{h'-a}{h'-x} = \frac{\frac{abc}{ab+bc+ca} - a}{\frac{abc}{ab+bc+ca} - \frac{a^2 b + a^2 c - abc}{a^2 - bc}}$$is pure imaginary, which follows by noting that $\overline{z} = -z$ (we see this by clearing denominators and expanding).

We $\sqrt{bc}$ invert and denote by $X'$ the inverse of $X$. Clearly $O'$ is the reflection of $A$ over $BC$, $H'$ is actually $A'$ (from the problem), $F'$ lies on $(BCO')$ such that $AO'=AF'$. Circle $(AFH)$ maps to $F'A'$, the circle with diameter $AA'$ maps to the line parallel to $BC$ through $H$ and the circle $(BOC)$ maps to the circle $(BCO')$. Let $T$ be the point on $A'F'$ such that $TH\parallel BC$. We wish to show that $HTF'O'$ is cyclic, or equivalently that $\angle TF'O'=90^\circ$. This is further equivalent to $AH$ and $F'A'$ intersecting on the circle with center $A$ and radius $AO'$. Looking at the diagram before the inversion, it would suffice to prove that the reflection of $O$ over $A$ lies on $(AHF)$. Denote this point by $E$. Now we proceed with complex numbers. Let $(ABC)$ be the unit circle. Clearly $e=2a$ and $ h=a+b+c$. \[\left\vert a-f\right\vert=1\iff \overline f=\frac{f}{af-a^2}\]\[\frac{b-f}{c-f}\cdot\frac{c}{b}\in\mathbb R\iff\frac{bc-cf}{bc-bf}=\frac{1-b\overline{f}}{1-c\overline{f}}\]\[\iff f\overline{f}(b-c)(b+c)-f(b-c)-bc\overline{f}(b-c)=0\iff \overline{f}=\frac{f}{f(b+c)-bc}\]Therefore, $f=\frac{a^2-bc}{a-b-c}$. Now we just need to check that \[\frac{a}{b+c}\cdot\frac{a+b+c-\frac{a^2-bc}{a-b-c}}{2a-\frac{a^2-bc}{a-b-c}}\in\mathbb R\iff \frac{ab^2+ac^2+abc}{(b+c)\left(a^2-2ab-2ac+bc\right)}=\frac{\frac{1}{a}}{\frac{1}{b}+\frac{1}{c}}\cdot\frac{\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{bc}}{\frac{1}{a^2}-\frac{2}{ab}-\frac{2}{ac}+\frac{1}{bc}}\]\[\iff \frac{ab^2+ac^2+abc}{a^2-2ab-2ac+b}=\frac{\frac{b}{c}+\frac{c}{b}+1}{\frac{1}{a}-\frac{2}{b}-\frac{2}{c}+\frac{a}{bc}}\]Which is clear. $\blacksquare$

Force overlay invert. Let $T,Z$ be the reflections of $A,O$ over $BC$ and let $S$ be the reflection of $T$ over $AZ$. The problem becomes showing that $(BHC)$, line $A'S$ and the line through $H$ parallel to $BC$ concur. We claim the point is the antipode $X$ of $T$ on $(BHC)$. Clearly $X\in(BHC)$ and $XH\perp HT\perp BC$. Thus we want $X,A',S$ collinear, and since $XS\perp ST\perp AZ$ this is the same as $AZ\parallel A'S$, or $[AZA']=[AZS]=-[AZT]=-[AOT]$. Now we get that $T=b+c-\frac{bc}a$ and $A'=\frac{bc+a^2}{b+c}$ so computation gives \[\begin{vmatrix}1&a&\frac1a\\1&0&0\\1&b+c-\frac{bc}a&\frac1b+\frac1c-\frac a{bc}\end{vmatrix}=\frac ba+\frac ca-\frac ab-\frac ac+\frac{a^2}{bc}-\frac{bc}{a^2}=-\begin{vmatrix}1&a&\frac1a\\1&b+c&\frac1b+\frac1c\\1&\frac{bc+a^2}{b+c}&\frac{bc+a^2}{a^2(b+c)}\end{vmatrix}\]as desired.

We will $\sqrt{bc}$ invert to get rid of the weird circle $(AFH)$; also it seems that this approach after the inversion has not been posted above. After the inversion, if $A_1$ is the reflection of $A$ in $BC$, $P \in (BHC)$ is such that $HP \parallel BC$ and $Q\in (BHC)$ is such that $AA_1=AQ$, then we have to show that $P, A', Q$ are collinear. We will show that $\angle A'PA_1=\angle QHA_1=\angle QPA_1$, which is sufficient. We first claim that $OPA'A_1$ is cyclic. Indeed, since $$\angle PA_1H=90^{\circ}-\angle HPA_1=90^{\circ}-\angle HCA_1=\beta-\gamma=\angle OAH,$$we have that if $T=PA_1 \cap AA'$, then $TA=TA_1$, i.e. $T \in BC$. Thus, by power of point, $TP \cdot TA_1=TB \cdot TC=TO \cdot TA'$, so $OPA'A_1$ is indeed cyclic, so we have shown $\angle A'PA_1=\angle A'OA_1$, and hence we need $\angle A'OA_1=\angle QHA_1$. If $\angle AA_1Q=\angle AQA_1=\varphi$, then $$\angle QHA_1=180^{\circ}-\angle HQA_1-\varphi=90^{\circ}+\beta-\gamma-\varphi$$and $$\angle A'OA_1=\angle OAH+\angle OA_1A=\beta-\gamma+\angle OA_1A,$$so we need $\angle OA_1A=90^{\circ}-\varphi$, which is equivalent to the circumcenter of $\triangle AA_1Q$ lying on $OA_1$. Indeed, if $OA_1 \cap BC=X$, we have to show that $X$ lies on the perpendicular bisector of $A_1Q$. But this perpendicular bisector is the line through $A$ and the circumcenter of $BHC$ and since this line is the reflection of $OA_1$ in $BC$, we obtain that $X$ lies on it, which finishes the problem.