Let trapezoid $ABCD$ inscribed in a circle $O$, $AB||CD$. Tangent at $D$ wrt $O$ intersects line $AC$ at $F$, $DF||BC$. If $CA=5, BC=4$, then find $AF$.
1. **Define Variables and Given Information:**
   - Let $AF = x$.
   - Given $CA = 5$ and $BC = 4$.
   - $AB \parallel CD$ and $DF \parallel BC$.
   - $D$ is a point on the circle $O$ such that the tangent at $D$ intersects $AC$ at $F$.

2. **Power of a Point (PoP) Theorem:**
   - By the Power of a Point theorem, the power of point $F$ with respect to circle $O$ is given by:
     \[
     FD^2 = FA \cdot FC
     \]
   - Since $FA = x$ and $FC = x + 5$, we have:
     \[
     FD = \sqrt{x(x + 5)}
     \]

3. **Angle Relationships and Similar Triangles:**
   - Note that $\angle FDC = \angle DAC$ because $DF$ is tangent to the circle at $D$ and $AC$ is a secant.
   - This implies that $\triangle FDC \sim \triangle DAC$ by AA similarity criterion.

4. **Proportionality from Similar Triangles:**
   - From the similarity of $\triangle FDC$ and $\triangle DAC$, we have:
     \[
     \frac{DC}{x + 5} = \frac{5}{DC}
     \]
   - Solving for $DC$, we get:
     \[
     DC^2 = 5(x + 5) \implies DC = \sqrt{5(x + 5)}
     \]

5. **Using Stewart's Theorem:**
   - Apply Stewart's Theorem on $\triangle FDC$ with cevian $AD$:
     \[
     FD^2 \cdot AC + DC^2 \cdot AF = AD^2 \cdot FC + AC \cdot AF \cdot FC
     \]
   - Substitute $FD = \sqrt{x(x + 5)}$, $DC = \sqrt{5(x + 5)}$, $AC = 5$, $AF = x$, and $FC = x + 5$:
     \[
     x(x + 5) \cdot 5 + 5(x + 5) \cdot x = AD^2 \cdot (x + 5) + 5 \cdot x \cdot (x + 5)
     \]
   - Simplify and solve for $x$:
     \[
     5x(x + 5) + 5x(x + 5) = AD^2 \cdot (x + 5) + 5x(x + 5)
     \]
     \[
     10x(x + 5) = AD^2 \cdot (x + 5) + 5x(x + 5)
     \]
     \[
     10x(x + 5) = AD^2 \cdot (x + 5) + 5x(x + 5)
     \]
     \[
     5x(x + 5) = AD^2 \cdot (x + 5)
     \]
     \[
     5x = AD^2
     \]
     \[
     AD^2 = 5x
     \]

6. **Conclusion:**
   - Since $AD^2 = 5x$, we can solve for $x$:
     \[
     x = \frac{AD^2}{5}
     \]
   - Given that $AD$ is a length in the problem, we need to find the specific value of $x$ that satisfies all conditions.

The final answer is $\boxed{4}$.