4. (42nd IMO Shortlist) Let $\triangle ABC$ be an acute-angled triangle. Construct isosceles triangles $\triangle DAC$, $\triangle EAB$, and $\triangle FBC$ outside $\triangle ABC$ such that $DA = DC$, $EA = EB$, $FB = FC$, $\angle ADC = 2 \angle BAC$, $\angle BEA = 2 \angle ABC$, and $\angle CFB = 2 \angle ACB$. Let $D'$ be the intersection of line $DB$ and $EF$, $E'$ be the intersection of line $EC$ and $DF$, and $F'$ be the intersection of line $FA$ and $DE$. Find the value of $\frac{DB}{DD'} + \frac{EC}{EE'} + \frac{FA}{FF'}$.
4. Since $\triangle A B C$ is an acute triangle

Therefore, $\angle A D C, \angle B E A, \angle C F B < \pi$
So $\angle D A C = \frac{\pi}{2} - \frac{1}{2} \angle A D C = \frac{\pi}{2} - \angle B A C$
$$
\angle B A E = \frac{\pi}{2} - \frac{1}{2} \angle B E A = \frac{\pi}{2} - \angle A B C
$$

So $\angle D A E = \angle D A C + \angle B A C + \angle B A E = \pi - \angle A B C < \pi$
Similarly, $\angle E B F < \pi, \angle F C D < \pi$
Therefore, hexagon $A B C D E F$ is a convex hexagon, and
$$
\angle A D C + \angle B E A + \angle C F B = 2(\angle B A C + \angle A B C + \angle A C B) = 2 \pi
$$

Let $\omega_{1}, \omega_{2}, \omega_{3}$ be the circles with centers at $D, E, F$ and radii $D A, E B, C F$ respectively.
Since $\angle A D C + \angle B E A + \angle C F B = 2 \pi$
Therefore, the circles $\omega_{1}, \omega_{2}, \omega_{3}$ intersect at a common point.
Let this intersection point be $O$, then point $O$ is the reflection of $C$ over the line $D F$, and also the reflection of $A$ over the line $D E$ and the reflection of $B$ over the line $E F$.
So $\frac{D B}{D D^{\prime}} = \frac{D D^{\prime} + D^{\prime} B}{D D^{\prime}} = 1 + \frac{D^{\prime} B}{D D^{\prime}} = 1 + \frac{S_{\triangle E F B}}{S_{\triangle E F D}} = 1 + \frac{S_{\triangle E F O}}{S_{\triangle E F D}}$
Similarly, $\frac{E C}{E E^{\prime}} = 1 + \frac{S_{\triangle D F O}}{S_{\triangle E F D}}, \frac{F A}{F F^{\prime}} = 1 + \frac{S_{\triangle D E O}}{S_{\triangle E F D}}$
Therefore, $\frac{D B}{D D^{\prime}} + \frac{E C}{E E^{\prime}} + \frac{F A}{F F^{\prime}} = 3 + \frac{S_{\triangle E F O} + S_{\triangle D F O} + S_{\triangle D E O}}{S_{\triangle E F D}} = 4$