how to bash this easily

We use the following key lemma: Lemma: Let $ABC$ have incenter $I$, and let the incircle meet $BC$ at $D$. Let the Feuerbach point be $Fe$, the reflection of $D$ over $I$ be $D'$, and the midpoint of $AI$ be $A'$. Then $A',D',Fe$ are collinear. Proof: Define $B',C',E',F'$ similarly. Note that $DB'C'\sim DE'F'$ from simple angle chasing, so by spiral similarity, $B'E',C'F'$ meet at the other intersection of $(DE'F')$ and $(DB'C')$. But $(DE'F')$ is the incircle, and $(DB'C')$ is the nine point circle of $BIC$. It is known that $Fe$ is the Poncelet point of $ABCI$, which means it lies on the nine point circle of $BIC$. Thus, $Fe$ is the intersection of $(DE'F')$ and $(DB'C')$, so $B',E',Fe$ are collinear. By symmetry, $A',D',Fe$ and $C',F',Fe$ are collinear. $\square$ Note that the above proof also works if we replace the incircle with an excircle. Now we return to the original problem. Let the center of $\Omega$ be $O$, let the tangents at $A$ and $B$ meet at $T_{AB}$, and define similarly for the other pairs of points. Let $M_{AB}$ be the midpoint of $OT_{AB}$, and define similarly for other pairs of points. Let the point diametrically opposite $A$ be $A'$, and define $B',C',D'$ similarly. Applying our lemma to $T_{CD}T_{AC}T_{AD}$, we get that $M_{CD},A',F_B$ are collinear. Similarly, $M_{CD},B',F_A$ are collinear. Now applying Pascal to $A'F_BBB'F_AA$, we get $M_{CD},O,AF_A\cap BF_B$ are collinear. But $A'B'$ is the pole of $AF_A\cap BF_B$ wrt $\Omega$, so $A'B'\perp OM_{CD}\perp CD$. Thus, $A'B'\parallel CD$. Similar statements hold for every other pair of sides. But then $\angle(A'B',B'C')=\angle(CD,AD)=\angle(BC,AB)$, so $\angle A'B'C'=\angle ABC$, and since similar identities hold elsewhere, we get $A'B'C'D'$ is similar to $ABCD$. $\square$

$\textbf{LEMMA:}$ $ABC$ be a triangle with contact triangle $DEF$, incenter $I$ and incircle $\odot(I)$. Let $D^*$ be the $D-$ antipode in $\odot(I)$. Let $K$ be an arbitary point on $\odot(I)$ and $K^*$ be the $K-$ antipode in $I$. Let $Fe$ be the Feuerbach point of $\Delta ABC$ and $X$ be the midpoint of $AI$. Let $XK^*\cap\odot(I)=T$. Then $KFe\cap DT\in\overline{AI}$. Animate $K$ on $\odot(I)$, the mappings $K\mapsto KFe$ and $K\mapsto K^*\mapsto T\mapsto DT$ are homographies between pencil of lines through $D$ and $Fe$ and $KFe\equiv DT$ when $K\equiv D$. Thus by degenerate steiner conic theorem we have that $KFe\cap DT$ traces a line. From (Feuerbach point) we have that $XD^*,Fe$ are collinear. Now when $K\equiv D^*$ we have $KFe\cap DT=X$ and when $K\equiv IFe\cap\odot(I)$ we have $KFe\cap DT=I$. Thus $KFe\cap DT$ traces out the line $AI$. Relabel the original problem as follows $\textbf{RELABELLED PROBLEM:}$ $ABC$ be a triangle with contact triangle $DEF$ and $K\in\odot(I)$. The tangent at $K$ meets $BC,CA,AB$ at $X,Y,Z$. Let $\mathcal{F}_{\Delta AYZ}$ be the Feuerbach point of $\Delta AYZ$. Similaly define $\mathcal{F}_{\Delta CXY}$ , $\mathcal{F}_{\Delta BXZ}$ , $\mathcal{F}_{\Delta ABC}$. Let $KK\cap YZ=K^*$. Similarly define $D^*,E^*,F^*$. Then $\Delta DEF\cup K\stackrel{-}{\sim} \Delta D^*E^*F^*\cup K^*$. We now just restrict our attention to lines $DK$ and $E^*F^*$. Let $P$ be the midpoint of $XI$. Let $E',F'$ be the $E,F$ antipodes WRT $\odot(I)$. From (Feuerbach Point) we have $P,E',\mathcal{F}_{\Delta CXY}$ and $P,F',\mathcal{F}_{\Delta BXZ}$ are collinear. Applying our $\textbf{LEMMA}$ we have that $F\mathcal{F}_{\Delta CXY},E\mathcal{F}_{\Delta BXZ},XI$ are concurrent at a point $\mathcal{P}_{E^*F^*}$. Thus by La Hire we have that $E^*F^*$ is the polar of $\mathcal{P}_{E^*F^*}$ and hence we have $DK\parallel E^*F^*$. Using the same procedure we can have the analogous parallel lines. Now by a little angle chasing we can get $\Delta DEF\stackrel{-}{\sim}\Delta D^*E^*F^*$. Repeating the same procedure we get $\Delta EFK\stackrel{-}{\sim}\Delta E^*F^*K^*$. Thus summing up all we get $\Delta DEF\cup K\stackrel{-}{\sim} \Delta D^*E^*F^*\cup K^*$. $\quad\blacksquare$