6. In rectangle $A B C D$, it is known that $A D=12, A B$ $=5, P$ is any point on side $A D$, $P E \perp B D, P F$ $\perp A C, E, F$ are the feet of the perpendiculars, respectively. Then, $P E+P F=$ $\qquad$
$=.6 \cdot \frac{60}{13}$.
Draw $A G \perp B D$ at $G$.
$\because$ In an isosceles triangle, the sum of the distances from any point on the base to the two legs is equal to the altitude on the leg,
$$
\therefore P E+P F=A G \text {. }
$$

Given $A D=12, A B=5$, then $B D=13$.
$$
\therefore A G=\frac{12 \times 5}{13}=\frac{60}{13} \text {. }
$$

Therefore, $P E+P F=\frac{60}{13}$.