1. We shall use the standard notations for $ABC$, i.e., $\angle ABC = \beta$, $BC = a$, etc. We also write $s = \frac{a+b+c}{2}$ for the semiperimeter and $r$ for the inradius.

Let $MN$ intersect the altitude $AD$ (where $D$ lies on $BC$) at the point $L$. We have that $\angle BAD = 90^\circ - \beta$ and $\angle AML = \angle BMN = \frac{\beta}{2}$. (Since $BMN$ is an isosceles triangle with $\angle MBN = 180^\circ - \beta$.) It is known that $AM = s - b$, so by the Sine Law in the triangle $AML$ we have

$$\frac{AM}{\sin \angle ALM} = \frac{AL}{\sin \angle AML} \Longrightarrow \frac{s-b}{\sin \left(90^\circ + \frac{\beta}{2}\right)} = \frac{AL}{\sin \frac{\beta}{2}} \Longrightarrow AL = (s-b) \tan \frac{\beta}{2} = r.$$

Analogously, if $PQ$ intersects $AD$ at $L'$, then $AL' = r$. Therefore, $L$ and $L'$ coincide, and since $A_1 = MN \cap PQ$ by definition, we conclude that $L = L' = A_1$. In particular, we can now view the point $A_2$ as the point on the $A$-altitude such that $AA_2 = 2r$. Analogously, $B_2$ and $C_2$ lie on the $B$-altitude and $C$-altitude, respectively, and $BB_2 = CC_2 = 2r$.
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Now let $X$ be the reflection of $A$ on the midpoint of $BC$ and define $XYZ$ analogously. So $XYZ$ is the triangle whose midpoints of sides are $A$, $B$, and $C$. Let $J$ be the incenter of this triangle. As the triangles $XYZ$ and $ABC$ are similar with ratio 2, the inradius of $XYZ$ is equal to $2r$. So if $JJ_0$ is perpendicular to $YZ$ (with $J_0$ on $YZ$), then $AA_2$ and $JJ_0$ are parallel (both perpendicular to $YZ$) and equal, hence $AA_2JJ_0$ is a rectangle and in particular $A_2$ is the foot of the perpendicular from $J$ to the $A$-altitude of $ABC$. It follows that $A_2$, $B_2$, and $C_2$ lie on the circle $\omega$ with diameter $JH$.

Now we finish with a simple angle chasing. The circle $k$ gives $\angle A_2B_2C_2 = \angle A_2HC_2 = 180^\circ - \angle AHC = \angle ABC$; similarly for the angles at $A_2$ and $C_2$. The desired similarity follows.