Example 1 In the tetrahedron $A B C D$, it is known that
$$
A B=A C=A D=D B=5, B C=3, C D=4 \text {. }
$$

Then the volume of the tetrahedron is $\qquad$ .
Given that $\angle B C D=90^{\circ}$.
As shown in Figure 1, take the midpoint $E$ of $B D$ and connect $A E$ and $C E$.
By the properties of a right-angled triangle, we have
$$
B E=C E=D E \text {. }
$$
Since $A B=A C=A D=D B=5$, we have
$\triangle A B E \cong \triangle A C E \cong \triangle A D E$.
Thus, $A E \perp B D, A E \perp E C$.
Therefore, $A E \perp$ plane $B C D$, which means $A E$ is the height of plane $B C D$.
The calculation shows that
$$
\begin{array}{l}
V_{\text {tetrahedron }ABCD}=\frac{1}{3} S_{\triangle B C D} \cdot A E \\
=\frac{1}{3} \times 6 \times \frac{5 \sqrt{3}}{2}=5 \sqrt{3} .
\end{array}
$$