Example 3 Let the midpoint of side $AB$ of $\triangle ABC$ be $N$, $\angle A > \angle B$, and $D$ be a point on ray $AC$ such that $CD = BC$. $P$ is a point on ray $DN$ and is on the same side of line $BC$ as point $A$, satisfying $\angle PBC = \angle A$. $PC$ intersects $AB$ at point $E$, and $BC$ intersects $DP$ at point $T$. Find the value of the expression $\frac{BC}{TC} - \frac{EA}{EB}$.
Solution: As shown in the figure, extend $BP$ to intersect line $AC$ at $F$. Thus, $\triangle ACB \sim \triangle BCF$, which implies $AC \cdot CF = BC^2$, so $CF = \frac{BC^2}{AC}$.

Notice that line $DTN$ intersects $\triangle ABC$, applying Menelaus' theorem we get $\frac{AN}{NB} \cdot \frac{BT}{TC} \cdot \frac{CD}{DA} = 1$.
Thus, $\frac{BT}{TC} = \frac{DA}{CD} \cdot \frac{BN}{AN} = \frac{DA}{CD} = \frac{AC}{BC} + 1$,
so $\frac{BC}{TC} = \frac{BT}{TC} + 1 = \frac{AC}{BC} + 2$.
Similarly, by line $DNP$ intersecting $\triangle ABF$,
we get $\frac{FP}{PB} \cdot \frac{BN}{NA} \cdot \frac{AD}{DF} = 1$.
By line $CEP$ intersecting $\triangle ABF$, we get $\frac{FP}{PB} \cdot \frac{BE}{EA} \cdot \frac{AC}{CF} = 1$, so
$\frac{EA}{BE} = \frac{FP}{PB} \cdot \frac{AC}{CF} = \frac{AN}{BN} \cdot \frac{FD}{AD} \cdot \frac{AC}{CF} = \frac{FD}{AD} \cdot \frac{AC}{CF}$
$= \frac{\frac{BC^2}{AC} + BC}{AC + BC} \cdot \frac{AC}{\frac{BC^2}{AC}} = \frac{BC}{AC} \cdot \frac{AC^2}{BC^2} = \frac{AC}{BC}$, therefore,
$$
\frac{BC}{TC} - \frac{EA}{BE} = 2.
$$