6. Given that the incircle $\odot O$ of quadrilateral $ABCD$ touches $AB$, $BC$, $CD$, and $DA$ at points $E$, $F$, $G$, and $H$ respectively, and $AF$, $DF$ intersect $EG$ at points $M$, $N$. If $BF=CF=5$, $EG=6$, then the length of $MN$ is ( ).
(A) 3
(B) 4
( C) 5
(D) an indeterminate value
6. A.

As shown in Figure 3, connect $EF$, $GF$, and $HF$. Draw a line through point $A$ parallel to $EG$, intersecting lines $EF$ and $FH$ at points $P$ and $Q$ respectively. Let $HF$ intersect $EG$ at point $T$.

It is easy to prove that $EF = FG$.
Thus, $\angle P = \angle FEG = \angle EGF$
$= \angle BEF = \angle AEP$
$\Rightarrow AE = AP$.
Similarly, $AH = AQ$.
Since $AE = AH$, it follows that $AP = AQ$.
Because $PQ \parallel EG$, we have $ME = MT$.
Similarly, $NT = NG$.
Therefore, $MN = \frac{1}{2} EG = 3$.