6. As shown in Figure 1, let $P$ be a point inside the convex quadrilateral $A B C D$. Draw perpendiculars from $P$ to $A B$, $B C$, $C D$, and $D A$, with the feet of the perpendiculars being $E$, $F$, $G$, and $H$, respectively. Given that $A H=3$, $H D=4$, $D G=1$, $G C=5$, $C F=6$, $F B=4$, and $B E-A E=1$. Then the perimeter of quadrilateral $A B C D$ is $\qquad$
6.34.

By the Pythagorean theorem, we have
$$
\begin{array}{l}
P A^{2}-A H^{2}=P D^{2}-D H^{2}, \\
P D^{2}-D G^{2}=P C^{2}-C G^{2}, \\
P C^{2}-C F^{2}=P B^{2}-B F^{2}, \\
P B^{2}-B E^{2}=P A^{2}-A E^{2} .
\end{array}
$$

Adding the above four equations, we get
$$
\begin{array}{l}
A H^{2}+D G^{2}+C F^{2}+B E^{2} \\
=A E^{2}+B F^{2}+C G^{2}+D H^{2} \\
\Rightarrow 9+1+36+B E^{2}=A E^{2}+16+25+16 \\
\Rightarrow(B E-A E)(B E+A E)=11 .
\end{array}
$$

Since $B E-A E=1$, we have $B E+A E=11$, which means $A B=11$.
Therefore, the perimeter of quadrilateral $A B C D$ is 34.