6. $P$ is a point inside $\triangle A B C$, and three equal-length line segments $D E, F G, H I$ are parallel to sides $A B, B C$, and $C A$ respectively, and all pass through point $P$. Given that $A B=12, B C=8$, $C A=6$. Then $A I: I F: F B$ equals $(\quad)$.
(A) $1: 2: 3$
(B) $1: 4: 3$
(C) $1: 5: 3$
(D) $1: 3: 4$
6. (C).

Let $A I=x, I F=y, F B=z$. From $D E / / A B, F G / /$ $B C, I H / / A C$ and $A I P D$ and $F B E P$ are parallelograms, we get $D P=I A=x, P E=F B=z$. Therefore, $D E=F G=H I$ $=x+z$.
Also, from $F G / / B C, H I / / C A$, we have
$$
\frac{F G}{B C}=\frac{A F}{A B}, \frac{H I}{C A}=\frac{B I}{A B},
$$

which means
$$
\begin{array}{l}
\frac{x+z}{8}=\frac{x+y}{12} \text { by the property of proportion } \frac{x+12}{20}, \\
\frac{x+z}{6}=\frac{y+z}{12}-\text { by the property of proportion } \frac{z+12}{18} . \\
\therefore\left\{\begin{array}{l}
3 x+5 y=24, \\
3 x+2 z=12 .
\end{array}\right.
\end{array}
$$

Also, from $x+y+z=12$, we solve to get
$$
x=\frac{4}{3}=\frac{12}{9}, y=\frac{60}{9}, z=4=\frac{36}{9} .
$$

Thus, $A I: I F: F B=12: 60: 36=1: 5: 3$.