Let $H_A, H_B, H_C$ be the reflections of $H$ in sides $BC, CA, AB$, respectively. Observe that $\angle AA_1H=\angle A'HH_A=\angle A'H_AH$ proving that $A_1$ lies on $\odot(AA'H_A)$. Invert at $H$ mapping $A$ to $H_A$; clearly, $\overline{A'B'C'} \mapsto \odot(A_1B_1C_1)$, so the tangency follows.

Nice problem Let $l$ be tangent to $\odot ABC$ at $P$. Let $HP$ meet $\odot ABC$ again in $Q$. Let $D = \odot ABC \cap AH$. Let $\omega$ be the circle passing through $H$ and $Q$ tangent to $\odot ABC$. We shall prove that $\odot A_{1}B_{1}C_{1}$ is $\omega$. Angle chasing yields that $AA'A_{1}D$ is cyclic. So $ A'H.HA_{1} = AH.HD = PH.HQ$. So $PA'QA_{1}$ is cyclic. So $\angle A'PQ = \angle A'A_{1}Q = \angle HA_{1}Q$. Let $RQ$ be tangent to $\odot ABC$ such that $R$ and $A'$ lie in the same half plane formed by $PQ$. Then $\angle HA_{1}Q = \angle A'PQ = \angle RQP = \angle RQH$. This means that $QR$ is tangent to $\odot A_{1}HQ$ also and so $\odot A_{1}HQ$ is tangent to $\odot ABC$ $\implies$ $A_{1}$ lies on $\omega$. Similarly $B_{1} , C_{1}$ also lie on $\omega$. So $\odot A_{1}B_{1}C_{1}$ is $\omega$ as desired . Hence proved. Q.E.D.

Nice......

Very nice problem! Let $\ell$ be tangent to $(ABC)$ at $P$ and let $\omega$ be the $9$-point circle of $ABC$. Furthermore, if $A_2$ is the midpoint of $HA'$, $B_2$ is the midpoint of $HB'$, $C_2$ is the midpoint of $HC'$ and $D= AH \cap BC$. Since the homothety centered at $H$ with ratio $\frac{1}{2}$ maps $(ABC)$ to $\omega$, we have that the line $A_2B_2C_2$ is tangent to $\omega$ at the midpoint $Q$ of $HP$. Now, observe that since $A_2$ is the midpoint of $HA'$ and $\angle HDA'= 90º$, then $A_2D=A_2H$. Moreover, $\angle HA_2D= 180º- 2 \angle A_2H= 180º- 2\angle AHA_1= \angle HAA_1$, because $AH=AA_1$. This implies that $AA_1DA_2$ is cyclic. $(\star)$ Similarly, $BB_1EB_2$ and $CC_1FC_2$ are cyclic, where $E= BH \cap AC, F= CH \cap AB$. Now, consider the inversion $\Phi$ centered at $H$ with radius $\sqrt{-HA.HD}$, hence $\Phi(\omega)= (ABC)$, and from $(\star)$, we have that $\Phi(A_1)=A_2, \Phi(B_1)=B_2, \Phi(C_1)=C_2$, and since line $A_2B_2C_2$ is tangent to $\omega$, we have that its image is tangent to the image of $\omega$, which is $(ABC)$. Hence, $(A_1B_1C_1)$ is tangent to $(ABC)$, as desired. $\blacksquare$

Different and long solution. Let's first define new points. $D,E,F,K$ be feet of perpendiculars from $A,B,C,H$ to $BC,CA,AB,l$, respectively. $O$ be center of $(ABC)$ and $G$ be tangency point of $l$ to $(ABC)$. $T=GH\cap (ABC)$ and $O'=HK\cap (ADK)$. $H_J$ be $JH\cap (ABC)$ for$J\in\{A,B,C\}$. And let $\omega=(ABC)$. It is well known that $DH=DH_A$ and similarly others. Since $O'H\cdot O'K=AH\cdot HD=BH\cdot HE=CH\cdot HF$, $O'$ lies on $(BEK)$ and $(CFK)$ also. Claim 1: $H,A_1,B_1,C_1$ all lie on a circle with center $O'$. Proof: Let $AO'\cap HA'=X_A$. Since $\angle HDA'=\angle HKA'=90$, $(HDKA')$ is cyclic $\implies \angle X_AA'D=\angle HA'D=\angle HKD=\angle O'KD=\angle O'AD=\angle X_AAD \implies AX_ADA'$ is cyclic $\implies \angle AX_AA'=\angle ADA'=90 \implies O'A$ is perpendicular bisector of $A_1H \implies O'H=O'A_1$. Similarly, we get $O'B_1=O'H=O'C_1$ and the result follows $\square$. Claim 2: $T-O'-O$ are collinear. Proof: Since $O'H\perp l\perp OG$, we get $O'H\parallel OG$. So, it is enough to prove that $\frac{TG}{TH}=\frac{OG}{O'H}$. Let $\angle OGH=\angle GHK=\alpha, HG=a,OG=b$ and $\cos(\alpha)=c$. Then we get $Pow(H,\omega)=OG^2-OH^2=OG^2-(HG^2+OG^2-2HG\cdot OG\cdot \cos(\alpha))=2HG\cdot OG\cdot \cos(\alpha)-HG^2$. Then we have that $$\frac{TG}{TH}=1+\frac{HG}{TH}=1+\frac{HG^2}{Pow(H,\omega)}=\frac{2HG\cdot OG\cdot \cos(\alpha)}{Pow(H,\omega)}=\frac{2OG\cdot HG\cdot \cos(\alpha)}{HA\cdot HH_A}={2OG\cdot HK}{2HA\cdot HD}=\frac{OG\cdot HK}{HA\cdot HD}=\frac{OG\cdot HK}{HO'\cdot HK}=\frac{OG}{HO'}$$$\square$. Finishing: $\angle O'TH=\angle OTH=\angle OGH=\angle O'HT \implies O'T \implies T,H,A_1,B_1,C_1$ lie on a circle with center $O'$. Since $T-O'-O$ are collinear, we get that this circle tangents $\omega$ $\blacksquare$. 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an older source here by LeVietAn