Example 5 Given in $\square A B C D$, $E, F$ are points on sides $A B, A D$ respectively, and $E F$ intersects diagonal $A C$ at point $P$. If $\frac{A E}{E B}=\frac{a}{b}, \frac{A F}{F D}=\frac{m}{n}\left(a, b, m, n \in \mathbf{R}_{+}\right)$, then the value of $\frac{A P}{P C}$ is $(\quad$.
(A) $\frac{a m}{a m+b m}$
(B) $\frac{b n}{a n+b m}$
(C) $\frac{a m}{a m+a n+b m}$
(D) $\frac{b n}{a n+b m+b n}$
(2008, National Junior High School Mathematics Competition, Tianjin Preliminary)
Solution: As shown in Figure 6,  
connect $B D$ and $A C$  
intersecting at point $O$.  
Taking $A$ as the viewpoint, apply the Angle Subtended Theorem to points $E$,  
$P$, $F$ and $B$, $O$, $D$ respectively, we get  
$$
\begin{array}{l}
\frac{\sin (\alpha+\beta)}{A P}=\frac{\sin \alpha}{A E}+\frac{\sin \beta}{A F}, \\
\frac{\sin (\alpha+\beta)}{A O}=\frac{\sin \alpha}{A B}+\frac{\sin \beta}{A D} .
\end{array}
$$

In $\triangle A D C$, by the Law of Sines, we have  
$$
\frac{\sin \alpha}{A B}=\frac{\sin \beta}{A D} \text {. }
$$

Substituting equation (3) into equation (2), we get  
$$
\frac{\sin (\alpha+\beta)}{A O}=\frac{2 \sin \alpha}{A B}=\frac{2 \sin \beta}{A D} \text {. }
$$
From equation (1), we get $\frac{A O}{A P}=\frac{\frac{\sin \alpha}{A E}}{\frac{2 \sin \alpha}{A B}}+\frac{\frac{\sin \beta}{A F}}{\frac{2 \sin \beta}{A D}}$, which is  
$$
\frac{A C}{2 A P}=\frac{A B}{2 A E}+\frac{A D}{2 A F} \text {. }
$$

Thus, $\frac{A B}{A E}+\frac{A D}{A F}=\frac{A C}{A P}$.  
Since $\frac{A E}{E B}=\frac{a}{b}, \frac{A F}{A D}=\frac{m}{n}$, we have  
$$
\frac{a+b}{a}+\frac{m+n}{m}=\frac{A C}{A P},
$$

which is $\frac{2 a m+a n+b m}{a m}=\frac{A C}{A P}$.  
Taking the reciprocal of equation (5) and applying the property of ratios, we get  
$$
\frac{a m}{(2 a m+a n+b m)-a m}=\frac{A P}{A C-A P},
$$

which is $\frac{A P}{P C}=\frac{a m}{a m+a n+b m}$.