6. As shown in Figure 1, in $\square A B C D$, the circle passing through points $A, B, C$ intersects $A D$ at point $E$, and is tangent to $C D$. If $A B=4, B E=5$, then the length of $D E$ is ( ).
(A) 3
(B) 4
(C) $\frac{15}{4}$
(D) $\frac{16}{5}$
Figure 1
6. (D).

As shown in Figure 4, connect $AC$ and $CE$.
Since $AE // BC$, quadrilateral $ABCE$ is an isosceles trapezoid.
Therefore, $AC = BE = 5$.
Also, because $DC // AB$ and $DC$ is tangent to the circle,
we have $\angle BAC = \angle ACD = \angle ABC$.
Thus, $AC = BC = AD = 5$, and $DC = AB = 4$.
Since $DC^2 = AD \cdot DE$, we get $DE = \frac{DC^2}{AD} = \frac{16}{5}$.