The triangle $ABC$ is $| BC | = a$ and $| AC | = b$. On the ray starting from vertex $C$ and passing the midpoint of side $AB$ , choose any point $D$ other than vertex $C$. Let $K$ and $L$ be the projections of $D$ on the lines $AC$ and $BC$, respectively, $K$ and $L$. Find the ratio $| DK | : | DL |$.
1. **Identify the given elements and setup the problem:**
   - We have a triangle \(ABC\) with \( |BC| = a \) and \( |AC| = b \).
   - Point \(D\) is chosen on the ray starting from vertex \(C\) and passing through the midpoint of side \(AB\).
   - \(K\) and \(L\) are the projections of \(D\) on the lines \(AC\) and \(BC\), respectively.

2. **Use the Law of Sines in triangles \( \triangle ACD \) and \( \triangle BCD \):**
   - Let \(\angle ACD = \alpha\) and \(\angle BCD = \beta\).
   - By the Law of Sines in \(\triangle ACD\):
     \[
     \frac{AD}{\sin(\angle ACD)} = \frac{CD}{\sin(\angle CAD)}
     \]
   - By the Law of Sines in \(\triangle BCD\):
     \[
     \frac{BD}{\sin(\angle BCD)} = \frac{CD}{\sin(\angle CBD)}
     \]

3. **Relate the angles using the projections \(K\) and \(L\):**
   - Since \(K\) and \(L\) are the projections of \(D\) on \(AC\) and \(BC\), respectively, we have:
     \[
     DK = D \sin(\angle ACD) \quad \text{and} \quad DL = D \sin(\angle BCD)
     \]

4. **Express the ratio \( \frac{DK}{DL} \):**
   - Using the expressions for \(DK\) and \(DL\):
     \[
     \frac{DK}{DL} = \frac{D \sin(\angle ACD)}{D \sin(\angle BCD)} = \frac{\sin(\angle ACD)}{\sin(\angle BCD)}
     \]

5. **Use the Law of Sines to relate \(\sin(\angle ACD)\) and \(\sin(\angle BCD)\):**
   - From the Law of Sines in \(\triangle ACD\) and \(\triangle BCD\):
     \[
     a \sin(\angle BCD) = b \sin(\angle ACD)
     \]
   - Therefore:
     \[
     \frac{\sin(\angle ACD)}{\sin(\angle BCD)} = \frac{a}{b}
     \]

6. **Conclude the ratio \( \frac{DK}{DL} \):**
   - Substituting the above result:
     \[
     \frac{DK}{DL} = \frac{a}{b}
     \]

The final answer is \( \boxed{\frac{a}{b}} \).