[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -18.145825013519765, xmax = 38.53620650714312, ymin = -10.429917850596546, ymax = 23.460905173507374; /* image dimensions */ /* draw figures */ draw((5.0130937395214925,13.662649486922337)--(-0.5,-1.58), linewidth(0.4) + blue); draw((-0.5,-1.58)--(18.4099,-1.72114), linewidth(0.4) + blue); draw((18.4099,-1.72114)--(5.0130937395214925,13.662649486922337), linewidth(0.4) + blue); draw((-0.5,-1.58)--(10.184376819390716,7.72437369706628), linewidth(0.4) + blue); draw((1.6424475820849858,4.3434576955805815)--(18.4099,-1.72114), linewidth(0.4) + blue); draw((1.6424475820849858,4.3434576955805815)--(10.184376819390716,7.72437369706628), linewidth(0.4) + blue); draw((-3.9735958511073424,7.196086109835047)--(9.717344654058754,20.240716675458682), linewidth(0.4) + blue); draw((-3.9735958511073424,7.196086109835047)--(10.184376819390716,7.72437369706628), linewidth(0.4) + blue); draw((9.717344654058754,20.240716675458682)--(1.6424475820849858,4.3434576955805815), linewidth(0.4) + blue); draw(circle((4.973871599608791,8.407677519123352), 5.255118339163495), linewidth(0.4) + linetype("4 4") + red); draw(circle((6.852952801866909,3.6601538735235093), 5.255118339163495), linewidth(0.4) + linetype("4 4") + red); draw((1.6424475820849858,4.3434576955805815)--(6.81373066195421,-1.5948180942754766), linewidth(0.4) + blue); draw((6.81373066195421,-1.5948180942754766)--(10.184376819390716,7.72437369706628), linewidth(0.4) + blue); draw((5.0130937395214925,13.662649486922337)--(4.934649459696092,3.152705551324365), linewidth(0.4) + blue); draw((1.3488303274367524,12.212325273339477)--(1.6424475820849858,4.3434576955805815), linewidth(0.4) + blue); draw((1.3488303274367524,12.212325273339477)--(10.184376819390716,7.72437369706628), linewidth(0.4) + blue); draw((1.3488303274367524,12.212325273339477)--(6.81373066195421,-1.5948180942754766), linewidth(0.4) + blue); /* dots and labels */ dot((-0.5,-1.58),dotstyle); label("$B$", (-0.3494430543560244,-1.217237792734236), NE * labelscalefactor); dot((5.0130937395214925,13.662649486922337),dotstyle); label("$A$", (5.163365494324635,14.026232825094166), NE * labelscalefactor); dot((18.4099,-1.72114),dotstyle); label("$C$", (18.55690035756221,-1.365232653101502), NE * labelscalefactor); dot((4.934649459696092,3.152705551324365),linewidth(4pt) + dotstyle); label("$H$", (5.089368064141002,3.444600308834644), NE * labelscalefactor); dot((10.184376819390716,7.72437369706628),linewidth(4pt) + dotstyle); label("$H_{B}$", (10.343185607178945,8.03244098021989), NE * labelscalefactor); dot((1.6424475820849858,4.3434576955805815),linewidth(4pt) + dotstyle); label("$H_{C}$", (1.796482420969333,4.628559191772772), NE * labelscalefactor); dot((-3.9735958511073424,7.196086109835047),dotstyle); label("$B'$", (-3.827322272986776,7.551457684026277), NE * labelscalefactor); dot((9.717344654058754,20.240716675458682),dotstyle); label("$C'$", (9.862202310985332,20.612004111437503), NE * labelscalefactor); dot((3.227911529694873,7.4648016277396385),linewidth(4pt) + dotstyle); label("$H'$", (3.387427169917443,7.773449974577176), NE * labelscalefactor); dot((6.81373066195421,-1.5948180942754766),linewidth(4pt) + dotstyle); label("$D$", (6.976302533823644,-1.291235222917869), NE * labelscalefactor); dot((1.3488303274367524,12.212325273339477),linewidth(4pt) + dotstyle); label("$D'$", (1.500492700234801,12.509285506329688), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Nice geo, Assume that the circle $(AH_BH_C)$ touches $B'C'$ the reflection of $BC$ across $H_BH_C$ and let it touch $B'C'$ in point $D'$. When we reflect the circle across $H_BH_C$ we know that the new circle must touch $BC$, let it touch in point $D$. Now notice the following: $$\angle H_CDH_B = \angle H_CD'H_B = \angle H_CAH_B = \alpha$$$$\angle H_CHH_B = 180 - \alpha = \angle H_CH'H_B$$where $H'$ is the reflection of $H$ across $H_BH_C$. The two conditions layed out give us that $H_CH'H_BD$ is cyclic, impliy that $H$ is the orthocenter of $\triangle H_CH_BD$ and that $H'$ is the orthocenter of $\triangle H_CD'H_B$. This implies that $D',H',H,D$ all belong to a line. From here we get that $\angle H_CH_BD = \gamma$ and that $\angle H_BH_CD = \beta$, these two imply that $H_BD \parallel AB$ and that $H_CD \parallel AC$, this implies that $AH_B =DH_C$ and that $DH_B=AH_C$. From the angle conditions we have that $H_CDB \sim H_BCD \sim ACB$, which gives us the following relation: $$\frac{AC}{AB}=\frac{H_BC}{H_BD} = \frac{H_BC}{AH_C}$$which implies the following condtion $\frac{H_BC}{AC} = \frac{AH_C}{AB}$. Thus if the last condition holds, we have the tangency condition.

Let $D$ lie on $BC$ s.t. $AC//DH_C$. By the condition, we have $DH_B//AB$. This implies $(AH_BH_C)$ is symmetric to $(DH_BH_C)$ wrt $H_BH_C$. Since $AH\bot BC$, we have the tangent at $D$ wrt $(DH_BH_C)$ is parallel to $BC$, which implies $BC$ is tangent to $(DH_BH_C)$. So we're done.