5. In the isosceles trapezoid $A B C D$, $A B / / C D$, and $A B > C D$. Let the eccentricity of the hyperbola with foci at $A, B$ and passing through point $D$ be $e_{1}$, and the eccentricity of the ellipse with foci at $C, D$ and passing through point $A$ be $e_{2}$, then $e_{1} e_{2}=$
5. 1 .

As shown in Figure 3, let $AB = x, CD = y, AD = BC = z, AC = BD = u$, then
$$
e_{1} = \frac{AB}{BD - AD} = \frac{x}{u - z}, \quad e_{2} = \frac{CD}{AD + AC} = \frac{y}{u + z}.
$$

Therefore,
$$
e_{1} e_{2} = \frac{xy}{u^2 - z^2}.
$$

Draw $DH \perp AB$, with the foot of the perpendicular at $H$. By the symmetry of the isosceles trapezoid, we get
$$
\begin{array}{l}
u^2 - z^2 = BD^2 - AD^2 = BH^2 - AH^2 = (BH + AH)(BH - AH) \\
= AB \cdot CD = xy
\end{array}
$$

Therefore,
$$
e_{1} e_{2} = \frac{xy}{u^2 - z^2} = \frac{xy}{xy} = 1.
$$