Problem 1831

Solution:
Let $P = CI_{A} \cap k$ and $Q = CI_{B} \cap k$. First we prove that the circumcircle $k_{1}$ of $\triangle I_{A} I_{B} C$ is tangent to $k$ if and only if $I_{A} I_{B} \parallel PQ$.
Let $T$ be the point on the tangent line to $k$ at $C$, for which $\angle ACT = \angle ABC$. If $k$ and $k_{1}$ are tangent then $CT$ is their common tangent line and therefore

$\angle CQP = \angle TCP = \angle TCI_{A} = \angle CI_{B} I_{A}$,
i.e. $I_{A} I_{B} \parallel PQ$. Conversely, if $I_{A} I_{B} \parallel PQ$, then $\angle TCI_{A} = \angle TCP = \angle CQP = \angle CI_{B} I_{A}$, implying that the line $CT$ is tangent to $k_{1}$.

It remains to prove that $I_{A} I_{B} \parallel PQ \Longleftrightarrow \frac{AD}{BD} = \frac{AC + CD}{BC + CD}$.
Since $PI_{A} = PA = PD$ and $QI_{B} = QB = QD$, we have
$$\begin{aligned} I_{A} I_{B} \parallel PQ &\Longleftrightarrow \frac{CI_{A}}{CI_{B}} = \frac{AP}{BQ} \\ & \frac{AC + CD - AD}{2 \cos \frac{\angle ACD}{2}} \\ & \Longleftrightarrow \frac{BC + CD - BD}{2 \cos \frac{\angle BCD}{2}} = \frac{2R_{ABC} \sin \frac{\angle ACD}{2}}{2R_{ABC} \sin \frac{\angle BCD}{2}} \\ & \Longleftrightarrow \frac{AC + CD - AD}{BC + CD - BD} = \frac{\sin \angle ACD}{\sin \angle BCD} \\ & \Longleftrightarrow \frac{AC + CD - AD}{BC + CD - BD} = \frac{AD}{BD} \Longleftrightarrow \frac{AC + CD}{BC + CD} = \frac{AD}{BD} \end{aligned}$$