My proof. Let $(DEF)$ be unit circle. Then $$a=\frac{2ef}{e+f}, b=\frac{2fd}{f+d}, c=\frac{2de}{d+e}$$and $$\bar{a}=\frac{2}{e+f},\bar{b}=\frac{2}{f+d},\bar{c}=\frac{2}{d+e}.$$Since $AX\perp BC$ and $X$ lies on $BC$, we get $$x=\frac{-d^{2} + de + df + ef}{e + f},\bar{x}=\frac{d^{2} + de + df - ef}{d^{2}e + d^{2}f}.$$Similarly, we have $$y=\frac{-e^{2} + de + df + ef}{d + f},$$$$\bar{y}=\frac{e^{2} + de - df + ef}{de^{2} + e^{2}f},$$$$z=\frac{-f^{2} + de + df + ef}{d + e},$$and $$\bar{z}=\frac{f^{2} - de + df + ef}{df^{2} + ef^{2}}.$$Since $A',B',C'$ are reflections of $X,Y,Z$ in the lines $EF,FD,DE$, we get $$a'=\frac{(\bar{e}-\bar{f})x'+\bar{e}f-\bar{f}e}{\bar{e}-\bar{f}}=\frac{e^2f^2+f^2d^2+d^2e^2-def(e+f-d)}{d^2(e+f)},$$$$b'=\frac{e^2f^2+f^2d^2+d^2e^2-def(-e+f+d)}{e^2(f+d)},$$and $$c'=\frac{e^2f^2+f^2d^2+d^2e^2-def(e-f+d)}{f^2(d+e)}.$$Now, we can check $$\frac{a-b}{a-c}= \frac{d^{2}f^{2} - e^{2}f^{2}}{d^{2}e^{2} - e^{2}f^{2}}=\frac{a'-b'}{a'-c'}.$$This means triangles $ABC$ and $A'B'C'$ are similar.

The two triangles seem to be spirally similar with center of spiral similarity at the image of the incenter after inversion around the circumcircle.

a_507_bc wrote: The two triangles seem to be spirally similar with center of spiral similarity at the image of the incenter after inversion around the circumcircle. True and it's known as X36.

Bump for a synthetic solution.

This is insane.

Let $ I, O, H $ be the incenter, circumcenter, orthocenter of $ \triangle ABC $, respectively, and let $ I' $ be the image of $ I $ under the inversion WRT $ \odot (ABC) $. Let $ M $ be the midpoint of $ IH $ and let $ I_A $ be the $ \angle A $ - excenter WRT $ \triangle ABC $. Let $ T $ be the midpoint of $ AH $ and let $ J $ be the reflection of $ I $ in $ EF $. Since $ J $ is the incenter of $ \triangle AYZ$, it follows that $ J, H, X, I_A $ are concyclic, hence $ \triangle ATJ \sim \triangle AOI $. Let $ AI $ intersect $ \odot (ABC) $ again at $ N $ and let $ K $ be the reflection of $ A $ in $OI $. Note that the reflection point of $ T $ in $ EF $ lies on $ OI $ and $K, N, I' $ are collinear, \begin{align*} \measuredangle (NI', NM) &= \measuredangle (NK, I_AH) \\ &= \measuredangle (NK, NI_A) + \measuredangle (NI_A, I_AH) \\ &= \measuredangle (OI, OA) + \measuredangle (JX, XH) \\ &= \measuredangle (TX, TJ) + \measuredangle (JX, XH) \\ &= \measuredangle (JX, JT) \\ &= \measuredangle (II', IA'). \end{align*}Thus, by combining $ MN : A'I = HI_A : 2XJ = AO : AI = NI' : II' $, we get $ \triangle I'MN \sim \triangle I'A'I \Longrightarrow \triangle I'OA \sim \triangle I'MA' $. Similarly, $ \triangle I'OB \sim \triangle I'MB', \triangle I'OC \sim \triangle I'MC' $, so it is easy to see $ \triangle ABC \cup O \cup I' \sim \triangle A'B'C' \cup M \cup I' $. $ \blacksquare $