Solution:

Let $X, Y, Z$ be the points of tangency of the inscribed circle with the sides $AB, AC, BC$, respectively. We prove that the line $IZ$ is tangent to both circles.

We denote by $\alpha, \beta, \gamma$ the measures of the angles at $A, B, C$, respectively. We denote by $T$ a point on the line $ZI$, on the opposite side of $Z$ with respect to $I$. Proving that the line $ZI$ is tangent to the circumscribed circle of triangle $BID$ is equivalent to proving that $\angle DBI = \angle DIT$. We will show that both angles are equal to $\beta$.

Indeed, the quadrilateral $IXBZ$ is cyclic, since it has right angles at $X$ and $Z$. It follows that $\angle TIX = \angle ZBX = \beta$, since both are supplementary to $\angle XIZ$. On the other hand $\angle DBI = 2 \angle XBI$ (since $BX$ is the altitude with respect to the base, and hence also the bisector, in the isosceles triangle $DBI$) and $\angle XBI = \beta / 2$ since $BI$ is the bisector of triangle $ABC$.

In a completely analogous way one proves that $\angle ECI = \angle EIT = \gamma$.

Solution:

If the circumscribed circles of triangles $BID$ and $CIE$ are tangent, then the point of tangency is necessarily $I$. Denoting by $O_{1}$ and $O_{2}$ the circumcenters of $BID$ and $CIE$, the claim holds if and only if the points $O_{1}, I$ and $O_{2}$ are collinear.

We denote by $\alpha, \beta, \gamma$ the measures of the angles at $A, B, C$, respectively. Triangle $BDI$ is isosceles, so $O_{1}$ belongs to the axis of $DI$, that is, to the line $AB$. The segments $O_{1}I$ and $O_{1}B$ are equal since they are radii, hence triangle $O_{1}IB$ is isosceles. Moreover the line $BI$ is the bisector of the angle at $B$. It follows that
$$\angle O_{1}IB = \angle IBO_{1} = \angle IBC = \frac{\beta}{2}$$
and therefore the lines $BC$ and $O_{1}I$ are parallel since they form equal alternate interior angles with the transversal $IB$.

In an analogous way one proves that the line $O_{2}I$ is parallel to $BC$.

But then $O_{1}I$ and $O_{2}I$ are parallel lines passing through $I$, hence they coincide, that is, the points $O_{1}, I, O_{2}$ are collinear, from which the claim follows.