7. Let the altitudes $A D, B E, C F$ of the acute triangle $\triangle A B C$ intersect at $H$. If $B C=a, A C=b, A B=c$. Then the value of $A H \cdot A D+B H$ - $B E+C H \cdot C F$ 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) $\frac{2}{3}\left(a^{2}+b^{2}+c^{2}\right)$
7. From the given, we know that points $H, D, C, E$ are concyclic, therefore, we have
$$
\begin{array}{l}
A D \cdot A H=A C \cdot A E \\
=A C \cdot A B \\
\quad \cdot \cos \angle B A E \\
=\frac{1}{2}\left(A C^{2}+A B^{2}-B C^{2}\right)=\frac{1}{2}\left(b^{2}+c^{2}-a^{2}\right) .
\end{array}
$$

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