As shown in Figure 6, let the radii of $\odot O_{1}$ and $\odot O_{2}$ be $r_{1}$ and $r_{2}$, respectively. Draw a line through point $B$ parallel to $EF$ intersecting $\odot O_{1}$ and $\odot O_{2}$ at points $G$ and $H$, respectively.
Let $B R \cap P Q=Z, B G \cap E P=X, B H \cap F Q=Y$,
$P Q \cap E O_{1}=N_{1}, P Q \cap F O_{2}=N_{2}$,
$B G \cap E O_{1}=M_{1}, B H \cap F O_{2}=M_{2}$.
$E M_{1}=F M_{2}=a, E N_{1}=F N_{2}=b$.
It is easy to see that $E G=\sqrt{2 r_{1} a}, E P=\sqrt{2 r_{1} b}$,
$E X=\frac{a}{b} E P=\sqrt{\frac{a}{b}} \cdot \sqrt{2 r_{1} a}$.
Thus, $\frac{E X}{E G}=\sqrt{\frac{a}{b}}$.
By $\triangle E G X \backsim \triangle B P X$, we know $\frac{E X}{E G}=\frac{B X}{B P}=\sqrt{\frac{a}{b}}$.
Similarly, $\frac{B Y}{B Q}=\sqrt{\frac{a}{b}}$.
Therefore, $\frac{B X}{B P}=\frac{B Y}{B Q}$, which means $\frac{B P}{B Q}=\frac{B X}{B Y}$.
Since $P Q \parallel X Y$, we have $\frac{P Z}{Q Z}=\frac{B X}{B Y}=\frac{B P}{B Q}$.
Thus, $\angle P B R=\angle Q B R$.