The quadrilateral $ABCD$  has an incircle of diameter $d$ which touches $BC$ at $K$ and touches $DA$ at $L$. Is it always true that the harmonic mean of $AB$ and $CD$ is equal to $KL$ if and only if the geometric mean of $AB$ and $CD$ is equal to $d$?
1. **Define the harmonic mean and geometric mean:**
   The harmonic mean \( h \) of \( AB \) and \( CD \) is given by:
   \[
   h = \frac{2}{\frac{1}{AB} + \frac{1}{CD}} = \frac{2AB \cdot CD}{AB + CD}
   \]
   The geometric mean \( g \) of \( AB \) and \( CD \) is given by:
   \[
   g = \sqrt{AB \cdot CD}
   \]

2. **Express the harmonic mean in terms of the geometric mean:**
   Let \( s = AB + CD \). Then, we can rewrite the harmonic mean \( h \) as:
   \[
   h = \frac{2AB \cdot CD}{AB + CD} = \frac{2g^2}{s}
   \]

3. **Use Pitot's theorem:**
   According to Pitot's theorem, for a quadrilateral with an incircle, the sum of the lengths of opposite sides is equal. Therefore, we have:
   \[
   AB + CD = BC + DA = s
   \]

4. **Relate \( KL \) to the diameter \( d \):**
   Given that the incircle touches \( BC \) at \( K \) and \( DA \) at \( L \), we need to show the relationship between \( KL \) and \( d \). Suppose:
   \[
   KL = \frac{2d^2}{s}
   \]

5. **Prove the equivalence:**
   - **If \( KL = h \):**
     \[
     KL = \frac{2d^2}{s} \implies \frac{2d^2}{s} = \frac{2g^2}{s} \implies d^2 = g^2 \implies d = g
     \]
     Therefore, \( (KL = h) \implies (g = d) \).

   - **If \( g = d \):**
     \[
     g = d \implies g^2 = d^2 \implies \frac{2g^2}{s} = \frac{2d^2}{s} \implies h = KL
     \]
     Therefore, \( (g = d) \implies (KL = h) \).

   - **Combining both implications:**
     \[
     (KL = h) \iff (g = d)
     \]

6. **Consider the case when \( KL \neq \frac{2d^2}{s} \):**
   - If \( KL \neq \frac{2d^2}{s} \), then:
     \[
     (KL \neq h) \implies \lnot(g = d)
     \]
   - If \( g \neq d \), then:
     \[
     (g \neq d) \implies \lnot(KL = h)
     \]

   Therefore, \( (KL = h) \land (g = d) = 0 \).

7. **Conclusion:**
   We have shown that \( (KL = h) \iff (g = d) \).

The final answer is \( \boxed{ (KL = h) \iff (g = d) } \).