1. Given Information and Initial Setup:
- Triangle $ABC$ is acute with $AC > BC$.
- $ABC$ is inscribed in a circle with center $O$.
- $CD$ is the diameter of the circle.
- Point $K$ is on the extension of ray $DA$ beyond $A$.
- Point $L$ is on segment $BD$ such that $DL > LB$.
- $\angle OKD = \angle BAC$ and $\angle OLD = \angle ABC$.

2. Angle Chasing and Congruence:
- Denote $\angle ACB = \alpha$ and $\angle ABC = \beta$.
- Let $N$ be the intersection of $CD$ with the circle centered at $O$ and radius $OK$.
- By angle chasing, we have $\angle NOD = \angle BOL = \angle COK = \beta - \alpha$.

3. Similarity of Triangles:
- Since $\angle ODB = \angle OBD = \alpha$ and $\angle ONK = \angle OKN = \alpha$, we have $\triangle OKN \cup \{D\} \sim \triangle ODB \cup \{L\}$.
- This similarity gives us the ratio $\frac{DL}{LB} = \frac{DK}{DN} = \frac{DK}{KC}$.

4. Application of Menelaus' Theorem:
- Let $M$ be the intersection of line $KL$ with segment $AB$.
- Applying Menelaus' theorem in $\triangle BCD$ with transversal $KML$, we get:
$$\frac{DL}{LB} \cdot \frac{BM}{MC} \cdot \frac{KC}{DK} = 1$$
- Using the ratio $\frac{DL}{LB} = \frac{DK}{KC}$, we substitute into the Menelaus' equation:
$$\frac{DK}{KC} \cdot \frac{BM}{MC} \cdot \frac{KC}{DK} = 1 \implies \frac{BM}{MC} = 1$$
- This implies $BM = MC$, meaning $M$ is the midpoint of $AB$.

Conclusion:
$$\boxed{\text{The line } KL \text{ passes through the midpoint of segment } AB.}$$