$7 \cdot 45$ In $\triangle A B C$, $A B=5, \quad B C=7, \quad A C=9, D$ is a point on $A C$ such that $B D=5$, then $A D: D C$ is
(A) $4: 3$.
(B) $7: 5$.
(C) $11: 6$.
(D) $13: 5$.
(E) $19: 8$.
[Solution] From the given and the cosine theorem in $\triangle B D C$ and $\triangle A B C$, we have
$$
B D^{2}=B C^{2}+C D^{2}-2 B C \cdot C D \cos \angle C \text {, }
$$

which is $C D^{2}-14 C D \cos \angle C+\left(7^{2}-5^{2}\right)=0$,
Similarly, $A C^{2}-14 A C \cos \angle C+\left(7^{2}-5^{2}\right)=0$.
From (1) and (2), we know that $C D, A C$ are the two roots of the equation
$$
x^{2}-14 \cos \angle C+24=0
$$

By Vieta's theorem, we know
$A C \cdot C D=24$, and since $A C=9$, then $C D=\frac{24}{9}=\frac{8}{3}$,
thus
$$
A D=9-\frac{8}{3}=\frac{19}{3} \text {. }
$$
$\therefore$
$A D: C D=19: 8$.

Therefore, the answer is $(E)$.