3. (10 points) From a point $P$ inside $\triangle A B C$, perpendiculars are drawn to the sides $B C, C A, A B$, with feet of the perpendiculars being $D, E, F$ respectively. Semi-circles are constructed outwardly on $A F, B F, B D, C D, C E, A E$ as diameters. As shown in the figure, the areas of these six semi-circles are denoted as $S_{1}, S_{2}, S_{3}, S_{4}, S_{5}, S_{6}$, respectively. If $S_{5}-S_{6}=2, S_{1}-S_{2}=1$, then $S_{4}-S_{3}=$ $\qquad$ .
3. (10 points) From a point $P$ inside $\triangle ABC$, perpendiculars are drawn to sides $BC$, $CA$, and $AB$, with feet of the perpendiculars being $D$, $E$, and $F$ respectively. Semi-circles are constructed outward with diameters $AF$, $BF$, $BD$, $CD$, $CE$, and $AE$. The areas of these six semi-circles are denoted as $S_{1}$, $S_{2}$, $S_{3}$, $S_{4}$, $S_{5}$, and $S_{6}$ respectively. Given that $S_{5}-S_{6}=2$ and $S_{1}-S_{2}=1$, find $S_{4}-S_{3}=3$.

【Analysis】Solution: As shown in the figure, connect $AP$, $BP$, and $CP$. According to the Pythagorean theorem, we have $AP^2 - BP^2 = AF^2 - BF^2$, $BP^2 - CP^2 = BD^2 - CD^2$, and $CP^2 - AP^2 = CE^2 - EA^2$. Therefore, $AF^2 + BD^2 + CE^2 = BF^2 + CD^2 + EA^2$. Then, using the formula for the area of a circle, we get $S_{1} + S_{3} + S_{5} = S_{2} + S_{4} + S_{6}$. Thus, $S_{4} - S_{3} = (S_{5} - S_{6}) + (S_{1} - S_{2}) = 2 + 1 = 3$. This completes the solution.

【Solution】Solution: As shown in the figure, connect $AP$, $BP$, and $CP$.

According to the Pythagorean theorem, we have:
$AP^2 = AF^2 + FP^2 \cdots$ (1),
$BP^2 = BF^2 + FP^2 \cdots$ (2),
(1) - (2), we get $AP^2 - BP^2 = AF^2 - BF^2 \cdots$ (3),

Similarly, we get $BP^2 - CP^2 = BD^2 - CD^2 \cdots$ (4),
Similarly, we get $CP^2 - AP^2 = CE^2 - EA^2 \cdots$ (5),
(3) + (4) + (5), we get $AF^2 + BD^2 + CE^2 = BF^2 + CD^2 + EA^2$,

Thus, $\frac{\pi}{4} \times (AF^2 + BD^2 + CE^2) = \frac{\pi}{4} \times (BF^2 + CD^2 + EA^2)$,
Therefore, $S_{1} + S_{3} + S_{5} = S_{2} + S_{4} + S_{6}$,
So, $S_{4} - S_{3}$
$$
\begin{array}{l}
= (S_{5} - S_{6}) + (S_{1} - S_{2}) \\
= 2 + 1 \\
= 3
\end{array}
$$

Hence, the answer is: 3.