Claim: $A_1,B_1,C_1$ lie on the respective heights of the triangle and $AA_1=BB_1=CC_1=r$. Proof: Let $T$ be the foot of the perpendicular from $A$ onto $BC$ and set $\{X\}=MN\cap AT$. Then from Menelaus's Theorem in $\triangle ABT$ with the secant $\overline{N,M,X}$ we have $$\frac{AM}{MB}\cdot \frac{BN}{NT}\cdot \frac{TX}{XA}=1.$$Simplifying using $AM=s-b$ and $BN=s-a$ yields $$AX=\frac{AT(s-b)}{c+BT}=\frac{c\sin B (s-b)}{c+c\cos B}=(s-b)\frac{\sin B}{1+\cos B}=(s-b)\tan \frac{B}{2}=r.$$Repeating the argument would give that $AY=r$, where $\{Y\}=PQ\cap AT$, meaning that $X=Y=A_1$, proving the first claim. Now set $D,E,F$ the points where the incircle $\omega$ touches the sides of $\triangle ABC$. $AA_2=2r$, so if $D'$ is the antipode of $D$ in $\omega$, $AA_2DD'$ would form a parallelogram. With complex numbers, this is $$a_2=2d+a=\frac{2(de+ef+fd)}{e+f},$$where we have chosen $\omega$ to be the unit circle. But now it's easy to see that $$\begin{vmatrix} \overline{a} & a_2 & 1\\ \overline{b} & b_2 & 1\\ \overline{c} & c_2 & 1\end{vmatrix} = 2(de+ef+fd)\begin{vmatrix} \frac{1}{e+f} & \frac{1}{e+f} & 1\\ \frac{1}{f+d} & \frac{1}{f+d} & 1\\ \frac{1}{d+e} & \frac{1}{d+e} & 1\end{vmatrix}=0,$$implying that triangles $\triangle ABC$ and $\triangle A_2B_2C_2$ are similar (and oppositely oriented).

Here is a synthetic solution after obtaining that $A_2, B_2, C_2$ are on the altitudes such that $AA_2=BB_2=CC_2=2r$. Let $XYZ$ be the triangle such that $A$ is midpoint of $YZ$, $B$ is midpoint of $XZ$ and $C$ - of $XY$. Let $J$ be its incenter, and let $X', Y', Z'$ be its touchpoints with the sides. Hence $JA_2AX'$ is a rectangle, so $\angle JA_2A=90$. If $H$ is the orthocenter, then $A_2, B_2, C_2$ lie on $(HJ)$, and we're done.

). Here's a motivation behind the proof (in particular constructing $J$ and showing all $A_2,B_2,C_2$ lie on $\odot(HJ)$): Claim 1: If $\triangle A_2B_2C_2 \sim \triangle ABC$, then $\triangle A_2B_2C_2$ and $\triangle ABC$ must be oppositely oriented. Proof: While this should be clear from a nice diagram, we provide a formal proof too. Assume on the contrary that $$ \triangle A_2B_2C_2 \stackrel{+}{\sim} \triangle ABC $$Due to mean geometry we obtain $$ \triangle A_1B_1C_1 \stackrel{+}{\sim} ABC$$Let $H = AA_1 \cap BB_1 \cap CC_1$ be orthocenter of $\triangle ABC$. Consider homothety $\mathcal T$ sending $A_1 \to A$. Let $\mathcal T(B_1) = B'$ and $\mathcal T(C_1) = C'$. Then we have, $$ \triangle ABC \stackrel{+}{\sim} \triangle A_1B_1C_1 \stackrel{+}{\sim} \triangle AB'C'$$So if $(B,C) \ne (B',C')$, then a spiral similarity at $A$ sends $BC \to B'C'$. As $BB' \cap CC' = H$, so it follows $$ A \in \odot(BHC) \implies \angle A = 90^\circ $$While if $(B,C) = (B',C')$, then $\triangle ABC$ and $\triangle A_1B_1C_1$ are homothetic. Combing this with $AA_1 = BB_1 = CC_1$ we obtain $$ HA = HB = HC $$We basically obtain that either $\angle A = 90^\circ$ or $HA = HB = HC$. But of course, this is not true for a general triangle. $\square$ Now from easy angle chase we have that $$ \triangle A_2B_2C_2 \stackrel{-}{\sim} \triangle ABC $$is equivalent to $$ H \in \odot(A_2B_2C_2)$$Let $A_0$ be the midpoint of arc $\widehat{BHC}$ of $\odot(BHC)$. Define $B_0,C_0$ similarly. Claim 2: $A_0 \in \odot(HB_2C_2)$. Proof: We only need $BB_2 = CC_2$ for this. Let $S$ be center of spiral similarity sending $BC \to B_2C_2$. Since spiral similarity comes in pairs, so $$ \triangle SBB_2 \sim \triangle SCC_2 $$Since $BB_2 = CC_2$, so the triangles are actually congruent, i.e. $$ \triangle SBB_2 \cong \triangle SCC_2 $$Hence $SB = SC$. Combining this with $S \in \odot(HBC)$, we obtain $S$ is a midpoint of arc $\widehat{BC}$ of $\odot(HBC)$. From configuration its clear that it must be midpoint lying on same side of $H$ as of line $BC$, i.e. $S \equiv A_0$. Since $S \in \odot(HB_2C_2)$ also, so our Claim is proven. $\square$ So if $\triangle A_2B_2C_2 \sim \triangle ABC$, then all seven points $$H,A_2,B_2,C_2,A_0,B_0,C_0$$must lie on a circle. If $I$ is the incenter of $\triangle ABC$, then $\odot(HA_0B_0C_0)$ is just the Hagge Circle circle of $I$ WRT $\triangle ABC$. Using properties of the Hagge Circle, we prove the following Lemma: Lemma 3: Let $XYZ$ be the Anticomplementary Triangle of $\triangle ABC$ (i.e. $\triangle ABC$ is Medial triangle of $\triangle XYZ$). . Let $P$ be any point, $Q$ be the isogonal conjugate of $P$ WRT $\triangle ABC$ and $\mathbb H_P$ be Hagge Circle of $P$ WRT $\triangle ABC$. Let $Q'$ be the point such that $$ ABC \cup \{Q\} \stackrel{+}{\sim} XYZ \cup \{Q'\} $$(basically $Q'$ is the point of $\triangle XYZ$ corresponding to the point $Q$ of $\triangle ABC$). Then $\mathbb H_P$ is circle with diameter $HQ'$. Proof: Let $H'$ be antipode of $H$ WRT $\mathbb H_P$ and $G$ be centroid of $\triangle ABC$. From the proof given here (the homothety thing at end), we obtain (in terms of complex numbers of vectors say) $$H' = 2 \cdot \frac{B+C}{2} - (2Q - A) = A+B+C - 2Q = 3G - 2Q$$Since homthety at $G$ with scale $-2$ sends $\triangle ABC \to \triangle XYZ$, thus $$ Q' = 3G- 2Q $$Hence $H' \equiv Q'$, as desired. $\square$ Now denote by $J$ the incenter of $\triangle XYZ$. Then by above, points $H,A_0,B_0,C_0$ must lie on circle with diameter $HJ$. Hence, $$ \triangle A_2B_2C_2 \triangle ABC \iff A_2,B_2,C_2 \in \odot(HJ)$$Now because of rectangles, we easily have $A_2,B_2,C_2 \in \odot(HJ)$, which completes the proof. $\blacksquare$