5. Let $\triangle A B C$ be an acute triangle, and construct isosceles triangles $\triangle D A C$, $\triangle E A B$, and $\triangle F B C$ outside $\triangle A B C$ such that $D A = D C$, $E A = E B$, and $F B = F C$, with $\angle A D C = 2 \angle B A C$, $\angle B E A = 2 \angle A B C$, and $\angle C F B = 2 \angle A C B$. Let $D^{\prime}$ be the intersection of line $D B$ and $E F$, $E^{\prime}$ be the intersection of $E C$ and $D F$, and $F^{\prime}$ be the intersection of $F A$ and $D E$. Find the value of $\frac{D B}{D D^{\prime}} + \frac{E C}{E E^{\prime}} + \frac{F A}{F F^{\prime}}$.
Solution: Since $\angle ABC$ is an acute triangle, then $\angle ADC$, $\angle BEA$, $\angle CFB < \pi$. Therefore,
$$
\begin{array}{l}
\angle DAC = \frac{\pi}{2} - \frac{1}{2} \angle ADC = \frac{\pi}{2} - \angle BAC. \\
\angle BAE = \frac{\pi}{2} - \frac{1}{2} \angle BEA = \frac{\pi}{2} - \angle ABC. \\
\text{Thus, } \angle DAE = \angle DAC + \angle BAC + \angle BAE \\
= \pi - \angle ABC < \pi. \\
\end{array}
$$

Similarly, $\angle EBF < \pi$, $\angle FCD < \pi$.
Therefore, the hexagon $DAEBCF$ is a convex hexagon, and
$$
\begin{array}{l}
\angle ADC + \angle BEA + \angle CFB \\
= 2(\angle BAC + \angle ABC + \angle ACB) = 2\pi.
\end{array}
$$

Let $w_1$, $w_2$, $w_3$ be the circles with centers at $D$, $E$, $F$ and radii $DA$, $EB$, $FC$ respectively. Since $\angle ADC + \angle BEA + \angle CFB = 2\pi$, these three circles intersect at a common point. Let this point be $O$, as shown in Figure 7. Then $O$ is the reflection of $C$ over $DF$.

Similarly, $O$ is also the reflection of $A$ over $DE$ and $B$ over $EF$. Therefore,
$$
\begin{array}{l}
\frac{DB}{DD'} = \frac{DD' + D'B}{DD'} = 1 + \frac{D'B}{DD'} \\
= 1 + \frac{S_{\triangle AKB}}{S_{\triangle FDF}} = 1 + \frac{S_{\triangle EOF}}{S_{\triangle DHF}}. \\
\end{array}
$$

Similarly,
$$
\begin{array}{l}
\frac{EC}{EE'} = 1 + \frac{S_{\triangle DFO}}{S_{\triangle DEF}} \cdot \frac{FA}{FF'} = 1 + \frac{S_{\triangle LDE}}{S_{\triangle UEF}}. \\
\text{Thus, } \frac{DB}{DD'} + \frac{EC}{EE'} + \frac{FA}{FF'} \\
= 3 + \frac{S_{\triangle OEF} + S_{\triangle ODF} + S_{\triangle UOF}}{S_{\triangle OHF}} = 4. \\
\end{array}
$$