Claim 1. The reflections $\ell_{a}, \ell_{b}$ and $\ell_{c}$ of the line $\ell$ in the lines $x, y$, and $z$, respectively, are concurrent at a point $T$ which belongs to $\omega$. ! Proof. Notice that $\angle\left(\ell_{b}, \ell_{c}\right)=\angle\left(\ell_{b}, \ell\right)+\angle\left(\ell, \ell_{c}\right)=2 \angle(y, \ell)+2 \angle(\ell, z)=2 \angle(y, z)$. But $y \perp B I$ and $z \perp C I$ implies $\angle(y, z)=\angle(B I, I C)$, so, since $2 \angle(B I, I C)=\angle(B A, A C)$, we obtain $$\angle\left(\ell_{b}, \ell_{c}\right)=\angle(B A, A C) .$$ Since $A$ is the reflection of $D$ in $x$, $A$ belongs to $\ell_{a}$; similarly, $B$ belongs to $\ell_{b}$. Then (1) shows that the common point $T^{\prime}$ of $\ell_{a}$ and $\ell_{b}$ lies on $\omega$; similarly, the common point $T^{\prime \prime}$ of $\ell_{c}$ and $\ell_{b}$ lies on $\omega$. If $B \notin \ell_{a}$ and $B \notin \ell_{c}$, then $T^{\prime}$ and $T^{\prime \prime}$ are the second point of intersection of $\ell_{b}$ and $\omega$, hence they coincide. Otherwise, if, say, $B \in \ell_{c}$, then $\ell_{c}=B C$, so $\angle(B A, A C)=\angle\left(\ell_{b}, \ell_{c}\right)=\angle\left(\ell_{b}, B C\right)$, which shows that $\ell_{b}$ is tangent at $B$ to $\omega$ and $T^{\prime}=T^{\prime \prime}=B$. So $T^{\prime}$ and $T^{\prime \prime}$ coincide in all the cases, and the conclusion of the claim follows. Now we prove that $X, X_{0}, T$ are collinear. Denote by $D_{b}$ and $D_{c}$ the reflections of the point $D$ in the lines $y$ and $z$, respectively. Then $D_{b}$ lies on $\ell_{b}, D_{c}$ lies on $\ell_{c}$, and $$\begin{aligned} \angle\left(D_{b} X, X D_{c}\right) & =\angle\left(D_{b} X, D X\right)+\angle\left(D X, X D_{c}\right)=2 \angle(y, D X)+2 \angle(D X, z)=2 \angle(y, z) \\ & =\angle(B A, A C)=\angle(B T, T C), \end{aligned}$$ hence the quadrilateral $X D_{b} T D_{c}$ is cyclic. Notice also that since $X D_{b}=X D=X D_{c}$, the points $D, D_{b}, D_{c}$ lie on a circle with centre $X$. Using in this circle the diameter $D_{c} D_{c}^{\prime}$ yields $\angle\left(D_{b} D_{c}, D_{c} X\right)=90^{\circ}+\angle\left(D_{b} D_{c}^{\prime}, D_{c}^{\prime} X\right)=90^{\circ}+\angle\left(D_{b} D, D D_{c}\right)$. Therefore, $$\begin{gathered} \angle\left(\ell_{b}, X T\right)=\angle\left(D_{b} T, X T\right)=\angle\left(D_{b} D_{c}, D_{c} X\right)=90^{\circ}+\angle\left(D_{b} D, D D_{c}\right) \\ =90^{\circ}+\angle(B I, I C)=\angle(B A, A I)=\angle\left(B A, A X_{0}\right)=\angle\left(B T, T X_{0}\right)=\angle\left(\ell_{b}, X_{0} T\right) \end{gathered}$$ so the points $X, X_{0}, T$ are collinear. By a similar argument, $Y, Y_{0}, T$ and $Z, Z_{0}, T$ are collinear. As mentioned in the preamble, the statement of the problem follows. Comment 1. After proving Claim 1 one may proceed in another way. As it was shown, the reflections of $\ell$ in the sidelines of $X Y Z$ are concurrent at $T$. Thus $\ell$ is the Steiner line of $T$ with respect to $\triangle X Y Z$ (that is the line containing the reflections $T_{a}, T_{b}, T_{c}$ of $T$ in the sidelines of $X Y Z$ ). The properties of the Steiner line imply that $T$ lies on $\Omega$, and $\ell$ passes through the orthocentre $H$ of the triangle $X Y Z$. ! Let $H_{a}, H_{b}$, and $H_{c}$ be the reflections of the point $H$ in the lines $x, y$, and $z$, respectively. Then the triangle $H_{a} H_{b} H_{c}$ is inscribed in $\Omega$ and homothetic to $A B C$ (by an easy angle chasing). Since $H_{a} \in \ell_{a}, H_{b} \in \ell_{b}$, and $H_{c} \in \ell_{c}$, the triangles $H_{a} H_{b} H_{c}$ and $A B C$ form a required pair of triangles $\Delta$ and $\delta$ mentioned in the preamble. Comment 2. The following observation shows how one may guess the description of the tangency point $T$ from Solution 1. Let us fix a direction and move the line $\ell$ parallel to this direction with constant speed. Then the points $D, E$, and $F$ are moving with constant speeds along the lines $A I, B I$, and $C I$, respectively. In this case $x, y$, and $z$ are moving with constant speeds, defining a family of homothetic triangles $X Y Z$ with a common centre of homothety $T$. Notice that the triangle $X_{0} Y_{0} Z_{0}$ belongs to this family (for $\ell$ passing through $I$ ). We may specify the location of $T$ considering the degenerate case when $x, y$, and $z$ are concurrent. In this degenerate case all the lines $x, y, z, \ell, \ell_{a}, \ell_{b}, \ell_{c}$ have a common point. Note that the lines $\ell_{a}, \ell_{b}, \ell_{c}$ remain constant as $\ell$ is moving (keeping its direction). Thus $T$ should be the common point of $\ell_{a}, \ell_{b}$, and $\ell_{c}$, lying on $\omega$.