Let us invert with respect to circle $(BCB_1C_1)$ (which has center $M$), where $P\mapsto P^*$. Since $C$ and $C_1$ are fixed, $\overline{CC_1}^*=(MCYXC_1)$. Therefore, we also have $\ell^*=(MUV)$. Now, by three tangents lemma, $MB_1$ and $MC_1$ are tangent to $(AB_1HC_1)$, and so this circles is orthogonal to $BCB_1C_1$ and thus remains fixed under inversion. Again by three tangents lemma, $\ell$ is tangent to $(AB_1HC_1)$, and therefore, by inverting, $\ell^*=(MUV)$ is tangent to $(AB_1HC_1)^*=(AB_1HC_1)$

Let $K$ be the second intersection point of $(ABC)$ and $(AB_1C_1)$, $J$ be the $A- Humpty$ point of $\triangle ABC$. By angle chansing, we have $K \in (MCC_1)$ and $KHUX$ is cyclic. Thus, $MU.MX=MH.MK=MJ.MA \Rightarrow XHJA$ is cyclic. Similarity, we have $YVJA$ is cyclic. Therefore, $MUJV$ is cyclic. Draw the tangent line $Jx$ of $(MUV)$, we have: $ \angle xJH=90^o -\angle xJM= 90^o -\angle JVM= 90^o - \angle JAY=\angle HAJ\Rightarrow Jx$ also tangents to $(AB_1C_1)$. Therefore, $(AB_1C_1)$ tangents to $(MCC_1)$

Here is another relatively elementary solution. Let $O'$ and $O_1$ be the circumcenters of triangles $MUV$ and $XMY$. First we prove that $\triangle XYM$ and $\triangle VUM$ are similar. Indeed, $\angle XMY = \angle YMX = \angle VMU$ and $\angle XYM = \angle XYC - \angle MYC = \angle AXM - \angle CC_{1}M = \angle XMB - \angle CC_{1}M = \angle MUV.$ Now we have $\angle XMO' = \angle YMO_1 = 90^{\circ} - \angle MXY$, hence $MO' \perp XY$ and $MO' \perp BC$. Since the perpendicular distances of $M$ to $UV$ and $XY$ are $\frac{BC_{1}}{2}$ and $AA_{1}$ respectively the coefficient of similarity is $\frac{BC_{1}}{2AA_{1}}$, hence we obtain: $$\frac{MO'}{MO_{1}} = \frac{BC_{1}}{2AA_{1}}.$$ Let $A_H$ be the midpoint of $AH$, which is the center of the circumcircle of $\triangle AB_{1}C_{1}$, since $\angle AB_{1}H = \angle AC_{1}H = 90^{\circ}$. This circle also passes through $H$. Let $N$ and $O$ be the intersection points of $AM$ with the circumcircle of $\triangle AB_{1}C_{1}$ and $MO'$ respectively. From $AH \perp BC$ and $MO \equiv MO' \perp BC$, we have $\angle NMO = \angle NAA_{H} = \angle ANA_{H} = \angle ONM = \phi$. From the similarity $\triangle MNO \sim \triangle ANA_{H}$, we have $$\frac{OM}{AA_{H}} = \frac{MN}{AN}.$$ Using $\triangle ANH \sim \triangle AA_{1}M$ we have $$\frac{AN}{AH} = \frac{AA_{1}}{AM}$$and after substituting $$\frac{OM}{AA_{H}} = \frac{MN}{AA_{1}} \cdot \frac{AM}{AH}. $$ Since $\angle A_{H}C_{1}M = \angle 90^{\circ} - \angle AC_{1}A_{H} + \angle CC_{1}M = 90^{\circ}$ the line $MC_{1}$ is tangent to $k(A_{H}, A_{H}A)$ and for the degree of $M$, we obtain $AM \cdot MN = MC_{1}^2$. Now we have the following: $$\frac{MO}{O_{1}M} = \frac{MO}{AA_{H}} \cdot \frac{AA_{H}}{O_{1}M} = \frac{MC_{1}^2}{AA_{1} \cdot AH} \cdot \frac{AH/2}{a/(4\cos{\beta})} = \frac{a \cos{\beta}}{2AA_{1}} \cdot \frac{a^2/4}{a^2/4} = \frac{BC_{1}}{2AA_{1}} = \frac{MO'}{O_{1}M}$$ which means that $MO = MO'$ so $O \equiv O'$. Now, since $O,N$ and $A_{H}$ are collinear the given circles are tangent at $N$.

Yeah, inversion kills this problem. Consider an inversion at $M$ fixing the circumcircle of $BC_1B_1C$. Then: $A$ and $D$ swap, where $D$ is the intersection of $MA$ and $(AB_1HC_1)$, since $MB_1,MC_1$ are tangents to $(ABHC_1)$. $X$ and $U$ swap as $CMC_1XY$ is cyclic and $M,U,X$ are collinear. $Y$ and $V$ swap as $CMC_1XY$ is cyclic and $M,V,Y$ are collinear. $\ell$ and $(UDVM)$ swap, where $X,A,Y\in\ell\parallel BC$. As $\ell$ is tangent to $(AB_1DHC_1)$ at $A$, then $(UDVM)$ is tangent to $(AB_1DHC_1)$ at $D$. We are done.

Too easy for a TST #6? Let $T$ be the $A-$humpty point. Note that $\angle MC_1U=\angle C_1CM=\angle C_1XM,$ and so triangles $C_XU$ and $XC_1M$ are similar, i.e. $MC_1^2=MU \cdot MX$. Thus, $MU \cdot MX=MC_1^2=MT \cdot MA,$ hence quadrilateral $AXUT$ is cyclic. Similarly, quadrilateral $AYVT$ is cyclic. Moreover, $MU \cdot MX=MT \cdot MA=MV \cdot MY,$ and so $XUVY$ is cyclic, too. Therefore, $\angle UTM=\angle YXM=\angle UVM,$ hence $T \in (MUV)$. Since $T \in (AB_1C_1)$, too, we are left to prove that $\angle HTM=\angle HAT+\angle TVM$, that is $\angle TVM=\angle AMB$, which is true since $\angle TVM=\angle YAM=\angle AMB$.

Notice $\angle C_{1}XM=\angle MYC$, $\angle C_{1}XM+\angle XMB=\angle C_{1}XA=180^{\circ}-\angle YCC_{1}=\angle MYC+\angle YVC$ So $\angle YVC=\angle XMB=\angle UXA$ Which means that $X,U,V,Y$is cyclic Consider that $\angle VCM=\angle MYC,\angle YMC=\angle YMC$ Hence $\triangle YMC\sim \triangle CMV$ So $MV\cdot MY=MC^{2}=MU\cdot MX$ Using $M$ as the inversion center and $MC$ as the inversion radius for inversion transformation So $U\mapsto X$,$V\mapsto Y$,$\odot (HC_{1}B_{1})\mapsto \odot (HC_{1}B_{1})$ Since $\angle YAC=\angle ACB=\angle AHB_{1}$ Which means that $XY$ is tangent to $\odot(HC_{1}B_{1})$ So $\odot(MUV)$ is tangent to $\odot(HC_{1}B_{1})$

Let $O_1$ be the circumcenter of $\triangle HB_1C_1$ , $O_2$ be the circumcenter of $\triangle MUV$ Define $K=\overline{AM}\cap(HB_1C_1)$ It's known that $MB_1,MC_1$ are tangent to $(HB_1C_1)$(by simple angle changing) So we have $MV \cdot MY = MC^2 = MB_1^2 = MK \cdot MA$ So $AKVY$ is cyclic,and we have that $\angle B_1XM = \angle B_1CM= \angle MB_1C$ So $MK \cdot MA = MB_1^2 = MU \cdot MX$ So $AXUK$ is cyclic So $K$ is the Miquel point of $XAYVMU$ So $KMXY$ is cyclic Let $MJ$ be the diameter of $(MUV)$ Then $\angle AKH=Rt\angle=\angle MKJ$ So $HKJ$ is collinear As $AH$ is perpendicular to $BC$,we have $MU\cdot MX=MY\cdot MV$ So $UVXY$ is cyclic So $\angle YMC=\angle XYM=\angle MUV$ So $BC$ is tangent to $(MUV)$,which by means $MJ$ is perpendicular to $BC$ So we get $\triangle AHK \sim \triangle MJK$ So $O_1,K,O_2$ is collinear ,which shows the result