Let $\triangle DEF$ be the orthic triangle of $\triangle ABC$. Then, as $BDHF$ is cyclic, we have $$AH \cdot h_a=AH \cdot AD=AF \cdot AB=(b \cos A)c=\frac{b^2+c^2-a^2}{2}$$where the last equality follows from cosine law. Similarly, we have $BH \cdot h_b=\frac{c^2+a^2-b^2}{2}$ and $CH \cdot h_c=\frac{a^2+b^2-c^2}{2}$. On adding these three equations, we get $$AH \cdot h_a+BH \cdot h_b+CH \cdot h_c=\frac{a^2+b^2+c^2}{2}$$