3. Let $O_{1}$ and $\rho$ be the center and radius of $k_{c}$. It is clear that $C, I, O_{1}$ are collinear and $C I / C O_{1}=r / \rho$. By Stewart's theorem applied to $\triangle O C O_{1}$,
$$O I^{2}=\frac{r}{\rho} O O_{1}^{2}+\left(1-\frac{r}{\rho}\right) O C^{2}-C I \cdot I O_{1} .$$
Since $O O_{1}=R-\rho, O C=R$ and by Euler's formula $O I^{2}=R^{2}-2 R r$, substituting these values in (1) gives $C I \cdot I O_{1}=r \rho$, or equivalently $C O_{1} \cdot I O_{1}=\rho^{2}=D O_{1}^{2}$. Hence the triangles $C O_{1} D$ and $D O_{1} I$ are similar, implying $\angle D I O_{1}=90^{\circ}$. Since $C D=C E$ and the line $C O_{1}$ bisects the segment $D E$, it follows that $I$ is the midpoint of $D E$. Second solution. Under the inversion with center $C$ and power $a b, k_{c}$ is transformed into the excircle of $\widehat{A} \widehat{B} C$ corresponding to $C$. Thus $C D=$ $\frac{a b}{s}$, where $s$ is the common semiperimeter of $\triangle A B C$ and $\triangle \widehat{A} \widehat{B} C$, and consequently the distance from $D$ to $B C$ is $\frac{a b}{s} \sin C=\frac{2 S_{A B C}}{s}=2 r$. The statement follows immediately. Third solution. We shall prove a stronger statement: Let $A B C D$ be a convex quadrilateral inscribed in a circle $k$, and $k^{\prime}$ the circle that is tangent to segments $B O, A O$ at $K, L$ respectively (where $O=B D \cap A C$ ), and internally to $k$ at $M$. Then $K L$ contains the incenters $I, J$ of $\triangle A B C$ and $\triangle A B D$. Let $K^{\prime}, K^{\prime \prime}, L^{\prime}, L^{\prime \prime}, N$ denote the midpoints of arcs $B C, B D, A C, A D, A B$ that don't contain $M ; X^{\prime}, X^{\prime \prime}$ the points on $k$ defined by $X^{\prime} N=N X^{\prime \prime}=$ $K^{\prime} K^{\prime \prime}=L^{\prime} L^{\prime \prime}$ (as oriented arcs); and set $S=A K^{\prime} \cap B L^{\prime \prime}, \bar{M}=N S \cap k$, $\bar{K}=K^{\prime \prime} M \cap B O, \bar{L}=L^{\prime} M \cap A O$. It is clear that $I=A K^{\prime} \cap B L^{\prime}, J=A K^{\prime \prime} \cap B L^{\prime \prime}$. Furthermore, $X^{\prime} \bar{M}$ contains $I$ (to see this, use the fact that for $A, B, C, D, E, F$ on $k$, lines $A D, B E, C F$ are concurrent if and only if $A B \cdot C D \cdot E F=B C \cdot D E \cdot F A$, and then express $A \bar{M} / \bar{M} B$ by applying this rule to $A M B K^{\prime} N L^{\prime \prime}$ and show that $A K^{\prime}, \bar{M} X^{\prime}, B L^{\prime}$ are concurrent). Analogously, $X^{\prime \prime} \bar{M}$ contains $J$. Now the points $B, \bar{K}, I, S, \bar{M}$ lie on a circle $(\angle B \overline{K M}=\angle B I \bar{M}=\angle B S \bar{M})$, and points $A, \bar{L}, J, S, \bar{M}$ do so as well. Lines $I \bar{K}, J \bar{L}$ are parallel to $K^{\prime \prime} L^{\prime}$ (because $\angle \overline{M K} I=\angle \bar{M} B I=$ $\left.\angle \bar{M} K^{\prime \prime} L^{\prime}\right)$. On the other hand, the quadrilateral $A B I J$ is cyclic, and simple calculation with angles shows that $I J$ is also parallel to $K^{\prime \prime} L^{\prime}$. Hence $\bar{K}, I, J, \bar{L}$ are collinear. !
Finally, $\bar{K} \equiv K, \bar{L} \equiv L$, and $\bar{M} \equiv M$ because the homothety centered at $M$ that maps $k^{\prime}$ to $k$ sends $K$ to $K^{\prime \prime}$ and $L$ to $L^{\prime}$ (thus $M, K, K^{\prime \prime}$, as well as $M, L, L^{\prime}$, must be collinear). As is seen now, the deciphered picture yields many other interesting properties. Thus, for example, $N, S, M$ are collinear, i.e., $\angle A M S=\angle B M S$. Fourth solution. We give an alternative proof of the more general statement in the third solution. Let $W$ be the foot of the perpendicular from $B$ to $A C$. We define $q=C W, h=B W, t=O L=O K, x=A L$, $\theta=\measuredangle W B O(\theta$ is negative if $\mathcal{B}(O, W, A), \theta=0$ if $W=O)$, and as usual, $a=B C, b=A C, c=A B$. Let $\alpha=\measuredangle K L C$ and $\beta=\measuredangle I L C$ (both angles must be acute). Our goal is to prove $\alpha=\beta$. We note that $90^{\circ}-\theta=2 \alpha$. One easily gets
$$\tan \alpha=\frac{\cos \theta}{1+\sin \theta}, \quad \tan \beta=\frac{\frac{2 S_{A B C}}{a+b+c}}{\frac{b+c-a}{2}-x}$$
Applying Casey's theorem to $A, B, C, k^{\prime}$, we get $A C \cdot B K+A L \cdot B C=$ $A B \cdot C L$, i.e., $b\left(\frac{h}{\cos \theta}-t\right)+x a=c(b-x)$. Using that $t=b-x-q-h \tan \theta$ we get
$$x=\frac{b(b+c-q)-b h\left(\frac{1}{\cos \theta}+\tan \theta\right)}{a+b+c} .$$
Plugging (2) into the second equation of (1) and using $b h=2 S_{A B C}$ and $c^{2}=b^{2}+a^{2}-2 b q$, we obtain $\tan \alpha=\tan \beta$, i.e., $\alpha=\beta$, which completes our proof.