4. As shown in the figure, $P A$ is tangent to $\odot O$ at point $A$, $P C$ intersects $\odot O$ at points $B$ and $C$, $P O$ intersects $\odot O$ at point $D$, and $A E \perp P O$ at point $E$. Connect $B E$ and extend it to intersect $\odot O$ at point $F$, then connect $O C$, $O F$, $A D$, and $A F$. If $\angle B C O=30^{\circ}$ and $\angle B F O=20^{\circ}$, then the measure of $\angle D A F$ is $\qquad$.
Connect $O A, O B, C D, C F$, then $O A \perp A P$ $\Rightarrow P A^{2}=P E \cdot P O=P B \cdot P C$,
thus $B, C, O, E$ are concyclic $\Rightarrow \angle B E P=30^{\circ}$.
Then $\angle B O E=10^{\circ} \Rightarrow \angle C O D=\angle F O D=130^{\circ}$,
hence $\angle C O F=100^{\circ} \Rightarrow \angle O C D=25^{\circ}$,
$\angle O C F=40^{\circ}$. Therefore $\angle D C F=65^{\circ} \Rightarrow \angle D A F=115^{\circ}$.