6. In the spatial quadrilateral $ABCD$, $AB \perp BC, BC$ $\perp CD, CD \perp DA, DA \perp AB$. If $BD=1$, then the maximum value of the area of $\triangle ABC$ is $\qquad$
6. $\frac{1}{4}$.

First, prove that quadrilateral $ABCD$ is a rectangle, which only requires proving that points $A, B, C, D$ are coplanar.
Use proof by contradiction.
As shown in Figure 1, assume point $D$ is outside the plane $ABC$, and draw $DD_{1} \perp$ plane $ABC$ at point $D_{1}$, and connect $AD_{1}, CD_{1}$. Then $CD_{1}$ is the projection of $CD$ in the plane $ABC$.
Since $CD \perp BC$, then $CD_{1} \perp BC$.
Similarly, $AD_{1} \perp AB$.
Thus, quadrilateral $ABCD_{1}$ is a rectangle.
Therefore, $AD^{2} + CD^{2} > AD_{1}^{2} + CD_{1}^{2} = AC^{2}$, which contradicts $CD \perp DA$.

Since the length of the diagonal of rectangle $ABCD$ is 1, the maximum area is achieved when it is a square, with a maximum area of $\frac{1}{2}$. Thus, the maximum area of $\triangle ABC$ is $\frac{1}{4}$.