Three. (25 points) As shown in Figure 3, $\triangle A B C$ is divided into six smaller triangles by three concurrent lines $A D$, $B E$, and $C F$. If the areas of $\triangle B P F$, $\triangle C P D$, and $\triangle A P E$ are all 1, find the areas of $\triangle A P F$, $\triangle D P B$, and $\triangle E P C$.
Three, let the areas of $\triangle A P F$, $\triangle D P B$, and $\triangle E P C$ be $x$, $y$, and $z$ respectively.
Since $\frac{S_{\triangle D P B}}{S_{\triangle D P C}}=\frac{B D}{D C}, \frac{S_{\triangle D A B}}{S_{\triangle A K C}}=\frac{B D}{D C}$,
then $\frac{S_{\triangle D P B}}{S_{\triangle D P C}}=\frac{S_{\triangle D X B}}{S_{\triangle A M C}}$, i.e., $\frac{y}{1}=\frac{x+y+1}{z+2}$.
Thus, $x-y=y z-1$.
Similarly, $y-z=z x-1, z-x=x y-1$.
Assume $x \geqslant y \geqslant z$, then
$$
y z-1=x-y \geqslant 0, x y-1=z-x \leqslant 0 \text {. }
$$

Therefore, $x y \leqslant y z$, i.e., $x \leqslant z$.
Thus, $x=y=z$.
From $0=x-y=y z-1=z^{2}-1$, we get $z=1$.
Therefore, $x=y=z=1$.
Hence, the areas of the three triangles are all 1.
(Respectfully, Nanjing University Affiliated High School, 210008)