6. $\odot O_{1}$ and $\odot O_{2}$ intersect at points $A$ and $B$, $CE$ and $DF$ are common tangents of the two intersecting circles, $C, D$ are on $\odot O_{1}$, $E, F$ are on $\odot O_{2}$, line $AB$ intersects $CE$ and $DF$ at $M$ and $N$ respectively. Then the result of $MN - CE$ is $(\quad)$.
(A) positive
(B) negative
(C) zero
(D) uncertain
6.A.

As shown in Figure 4, by the secant-tangent theorem, we have
$$
\begin{array}{l}
C M^{2}=M A \cdot M B \\
=E M^{2} .
\end{array}
$$

Thus, $C M=E M$.
Similarly, $D N=F N$.
Since $C E=D F$, we have
$$
C M=D N .
$$

Also, $D N^{2}=N B \cdot N A$, so
$M A \cdot M B=N B \cdot N A$
$\Rightarrow M A(M A+A B)=N B(N B+A B)$
$\Rightarrow(M A-N B)(M A+N B+A B)=0$
$\Rightarrow M A=N B$.
Therefore, $M N=M B+N B=M B+M A$
$$
>2 \sqrt{M B \cdot M A}=2 \sqrt{C M^{2}}
$$
$=2 C M=C E$ (here $M B \neq M A)$.
Hence, $M N-C E>0$.