6. As shown in Figure $2, C E$ and $C F$ bisect $\angle A C B$ and $\angle A C D$ respectively, $A E \parallel C F$, $A F \parallel C E$, line $E F$ intersects $A B$ and $A C$ at points $M$ and $N$. If $B C = a, A C = b, A B = c$, and $c > a > b$, then the length of $M E$ is ().
(A) $\frac{c-a}{2}$
(B) $\frac{a-b}{2}$
(C) $\frac{c-b}{2}$
(D) $\frac{a+b-c}{2}$
6.B.

Since $CE, CF$ bisect $\angle ACB, \angle ACD$ respectively, we have $\angle ECA = \frac{1}{2} \angle ACB, \angle ACF = \frac{1}{2} \angle ACD$. Therefore, $\angle ECF = \angle ECA + \angle ACF = \frac{1}{2} (\angle ACB + \angle ACD) = 90^\circ$. Since $AE \parallel CF, AF \parallel CE$, quadrilateral $AECF$ is a rectangle. Thus, $EN = CN = \frac{1}{2} AC = \frac{b}{2}$, and $\angle NEC = \angle NCE$.
Since $\angle NCE = \angle ECB$, then $\angle NEC = \angle ECB$, so $EN \parallel BC$.
Hence, $\triangle AMN \sim \triangle ABC$.
Therefore, $\frac{MN}{BC} = \frac{AN}{AC} = \frac{1}{2}, MN = \frac{1}{2} BC = \frac{1}{2} a$.
Thus, $ME = MN - EN = \frac{a}{2} - \frac{b}{2} = \frac{a - b}{2}$.