One. (50 points) As shown in Figure 1, given that trapezoid $ABCD$ is inscribed in $\odot O$, with $AB \parallel CD$. A tangent line is drawn through point $D$ intersecting the extension of $CA$ at point $F$, and $DF \parallel BC$. If $CA = 5, BC = 4$, find the length of segment $AF$.
 Extend $BA$ to intersect $DF$ at $E$, and connect $BD$, then
$$
\angle DAC = \angle DBC.
$$

Since $DF$ is tangent to $\odot O$ at point $D$, we have
$$
\angle ADF = \angle ACD.
$$

From $AB \parallel CD$, we know
$$
\overparen{AD} = \overparen{BC} \Rightarrow \angle BDC = \angle ACD.
$$

Thus, $\angle ADF = \angle BDC$.
Therefore, $\angle FDB = \angle ADC$.
By $DF \parallel BC$, we get $\angle FDB = \angle DBC$. So, $\angle ADC = \angle DAC$.
Hence, $DC = AC = 5$.
It is also easy to see that quadrilateral $DCBE$ is a parallelogram, so $BE = DC = 5$, $DE = BC = 4$.
By the secant-tangent theorem, we have $ED^2 = EA \cdot EB$, i.e., $4^2 = EA \cdot 5 \Rightarrow EA = \frac{16}{5}$.
By $\triangle AEF \sim \triangle CDF$, we get
$$
\frac{AE}{DC} = \frac{AF}{FC} \Rightarrow \frac{\frac{16}{5}}{5} = \frac{AF}{AF + 5} \Rightarrow AF = \frac{80}{9}.
$$