To prove that the points $I$, $K$, and $H$ are collinear in triangle $ABC$ with orthocenter $H$ and incenter $I$, given that the point $K$ is chosen such that:

$AK + AH = BK + BH = CK + CH,$

we'll approach the problem by using concepts from triangle geometry, specifically isogonal conjugates.

Step-by-Step Proof

1. Understanding the Condition: The given condition $AK + AH = BK + BH = CK + CH$ suggests that point $K$ is equidistant from each vertex $A$, $B$, and $C$ after accounting for the distances to the orthocenter $H$. This setup is reminiscent of a point associated with isogonal conjugates in triangle geometry.

2. Isogonal Conjugate of the Incenter: In any triangle, the isogonal conjugate of the orthocenter $H$ is known to be the incenter $I$. Similarly, the isogonal conjugate of the incenter $I$ is the exsimilicenter of the incircle and circumcircle of the triangle, denoted by $X$, which is often represented as a point on the line joining $I$ and $H$.

3. Characterizing Point $K$: From the given conditions, $K$ must satisfy a symmetry relative to the triangle's vertices and the orthocenter, which aligns with the definition of the exsimilicenter $X$ of the incircle and circumcircle of $\triangle ABC$. This point $X$ has a special property: it lies on the line $IH$ (the line joining the incenter $I$ and the orthocenter $H$).

4. Conclusion: Since $K$ meets the same conditions that define the exsimilicenter $X$ of the incircle and circumcircle, $K$ must be the exsimilicenter $X$. Thus, $K$ lies on the line joining $I$ and $H$.

Therefore, we conclude that the points $I$, $K$, and $H$ are collinear.

$\boxed{I, K, \text{ and } H \text{ are collinear.}}$

Summary

This proof relies on recognizing that the configuration described by the distances to the orthocenter corresponds to the exsimilicenter of the incircle and circumcircle of the triangle, which is known to lie on the line $IH$.