17・109 Let the altitudes $A D, B E, C F$ of an acute triangle $\triangle A B C$ intersect at $H$. If $B C=$ $a, A C=b, A B=c$, then the value of $A H \cdot A D+B H \cdot B E+C H \cdot C F$ is
(A) $\frac{1}{2}(a b+b c+c a)$.
(B) $\frac{1}{2}\left(a^{2}+b^{2}+c^{2}\right)$.
(C) $\frac{2}{3}(a b+b c+c a)$.
(D) $\frac{2}{3}\left(a^{2}+b^{2}+c^{2}\right)$.
(China Junior High School Mathematics League, 1994)
[Solution] From the given, we know that points $H, D, C, E$ are concyclic, thus we have
$$
\begin{aligned}
A D \cdot A H & =A C \cdot A E=A C \cdot A B \cdot \cos \angle B A E \\
& =\frac{1}{2}\left(A C^{2}+A B^{2}-B C^{2}\right) \\
& =\frac{1}{2}\left(b^{2}+c^{2}-a^{2}\right) .
\end{aligned}
$$

Similarly,
$$
\begin{array}{c}
\text { Similarly, } B H \cdot B E=\frac{1}{2}\left(c^{2}+a^{2}-b^{2}\right), \\
C H \cdot C F=\frac{1}{2}\left(a^{2}+b^{2}-c^{2}\right) . \\
\therefore A H \cdot A D+B H \cdot B E+C H \cdot C F=\frac{1}{2}\left(a^{2}+b^{2}+c^{2}\right) .
\end{array}
$$

Therefore, the answer is $(B)$.