Example 1 In $\triangle ABC$, if $AB=5, BC=6$, $CA=7, H$ is the orthocenter, then the length of $AH$ is $\qquad$
$(2000$, Shanghai Junior High School Mathematics Competition)
Extend $A H$, intersecting $B C$ at point $D$, and connect $B H$ and extend it to intersect $A C$ at point $E$.
By the properties of the orthocenter, we know
$A D \perp B C$,
$B E \perp A C$.
In Rt $\triangle A B E$ and Rt $\triangle C B E$, by the Pythagorean theorem, we get
$$
\begin{array}{l}
B E^{2}=A B^{2}-A E^{2}=25-A E^{2}, \\
B E^{2}=B C^{2}-C E^{2}=36-(7-A E)^{2} .
\end{array}
$$

Thus, $25-A E^{2}=36-(7-A E)^{2} \Rightarrow A E=\frac{19}{7}$.
Therefore, $B E=\sqrt{A B^{2}-A E^{2}}=\frac{12 \sqrt{6}}{7}$.
It is easy to see that $\triangle A H E \backsim \triangle B C E \Rightarrow \frac{A H}{B C}=\frac{A E}{B E}$.
So, $A H=\frac{A E \cdot B C}{B E}=\frac{19 \sqrt{6}}{12}$.