5. As shown in Figure 2, in parallelogram $A B C D$, $A B=15$, a circle is drawn through point $D$ that is tangent to $A B$ and $B C$, and intersects sides $A D$ and $C D$ at points $E$ and $F$, respectively, with $5 A E=4 D E$ and $8 C F=D F$. Then $B H$ equals ( ).
(A) 5
(B) 6
(C) 7
(D) 8
5. (C).
$\because 8 C F=D F, \therefore C F=15 \times \frac{1}{9}=\frac{5}{3}$.
By the secant-tangent theorem, we have $C H^{2}=C F \cdot C D$,
thus $C H=5$. Let $B C=x>0$.
$\because B H=x-5=B G, \therefore A G=20-x$.
Also, $\because 5 A E=4 D E, \therefore D E=\frac{5}{9} x, A E=\frac{4}{9} x$.
By the secant-tangent theorem, we have $A G^{2}=A E \cdot A D$. Thus, $(20-x)^{2}=\frac{4}{9} x^{2}$. Solving for $x$ gives $x=12$. Therefore, $B H=B C-C H=7$.