If $O_1,O_2$ - centers of $(BID), (CIE)$, then $\angle AO_2I=\angle C$ and $\angle AO_1I=\angle B$. Hence $IO_1\parallel BC \parallel IO_2$, i.e. $I\in O_1O_2$.

Let $F$ be the intouch point of the incircle to $BC$, let $(BID) \cap AB=G$ and $(CIE) \cap AC=J$. Note that $GB,JC$ are diameters on $(BID)$ and $(CIE)$ respectivily. By angle chase $\angle IGB=90-\angle ABI=90-\angle CBI=\angle BIF$ thus we have $IF$ tangent to $(BID)$ By more angle chase $\angle IJC=90-\angle ACI=90-\angle BCI=\angle CIF$ thus we have $IF$ tangent to $(CIE)$ And we are done becuase $(BID)$ and $(CIE)$ share common tangent thus these circles are tangent

Let AB BC CA touch incircle at X Y Z. We will show IY is tangent to both BID and CIE. ∠BIY = 90-∠B/2 = ∠BDX = ∠BDI so IY is tangent to BID. similarly we can prove that IY is tangent to CIE so circles are tangent themselves.

Let $\angle ABC=2\beta, \angle BCA=2\gamma, \angle CAB=2\alpha$. Let point $X$ lie on $BC$ such that $IX\perp BC$, $ID$ meet $AB$ at $Y$, $IE$ meet $AC$ at $Z$. Claim: $IX$ is tangent to $(BID)$ and $CIE$. Proof: In $\Delta IXC, \angle ICX=\gamma\implies\angle CIX=90^{\circ}-\gamma$. By properties of reflection, $IZ=ZE$, $IE\perp CZ \implies \Delta CIE$ is isosceles $\implies \angle ECZ=\angle ICZ=\gamma$. $\implies \angle CEI=90^{\circ}-\gamma=\angle CIX$. By the converse of alternate segment theorem, we get that $(CIE)$ is tangent to $IX$. In $\Delta IXB, \angle IBX=\beta\implies\angle BIX=90^{\circ}-\beta$. By properties of reflection, $IY=YD$, $ID\perp BY \implies \Delta BID$ is isosceles $\implies \angle DBY=\angle IBY=\beta$. $\implies \angle BDI=90^{\circ}-\beta=\angle BIX$. By the converse of alternate segment theorem, we get that $(BID)$ is tangent to $IX$. Therefore, as $IX$ is tangent to both the circles at point $X$, then the two circles $(CIX)$ and $(BIE)$ must also be mutually tangent to each other, and we're done!

Let $XYZ$ be the contact triangle of $ABC$. We invert about the incircle and let $\bullet^*$ denote the image of $\bullet$. Clearly, $(BID)$ and $(CIE)$ each map to $B^*D^*$ and $C^*E^*$. Moreover, it's easy to see that $B^*$ and $C^*$ are the midpoints of $ZX$ and $XY$, while $D^*$ and $E^*$ are the midpoints of $IZ$ and $IY$. Thus, $B^*D^*$ and $C^*E^*$ are just the $Y$-midline of $IXY$ and $Z$-midline of $ZIX$ respectively, whence $B^*D^* \parallel IX \parallel C^*E^*$. $\blacksquare$

It suffices to show that $$\angle DIE = \angle DBI + \angle ICE.$$which is true as the angles are respectively $\pi - A, B, C$.