【Question 10】
As shown in the figure, $\triangle ABC$ is an equilateral triangle. Take points $D, E$ on side $BC$ such that $BC=3DE$. Construct an equilateral $\triangle DEF$, and connect $AF$. Draw $DG$ parallel to $AF$ intersecting side $AB$ at point $G$, and draw $EH$ parallel to $AF$ intersecting side $AC$ at point $H$. Draw $GI \perp AF, HJ \perp AF$. If the area of $\triangle BDF$ is 45, and the area of $\triangle DEF$ is 30, then $GI \div HJ=$ $\qquad$
【Analysis and Solution】
Geometry, Equal Area Transformation, Swallowtail Model.
Extend $A F$ to intersect $B C$ at point $K$, and connect $A D, A E$.
By the common side model, we get $B D: D E=S_{\triangle B D F}: S_{\triangle D E F}=45: 30=3: 2$;
Since $B C=3 D E$;
Therefore, $B D: D E: B C=3: 2: 6$;
Thus, $B D: D E: C E=3: 2: 1$.
Because $\angle A B C=\angle F D E=60^{\circ}$;
Therefore, $D F \parallel A B$;
So, $D K: B K=D F: A B=1: 3$;
$D K: B D=1: 2$;

Because $\angle F E D=\angle A C B=60^{\circ}$;
Therefore, $E F \parallel A C$;
So, $E K: C K=E F: A C=1: 3$;
Thus, $E K: C E=1: 2$;
Since $B D: D E: C E=3: 2: 1$;
Therefore, $B D: D K: E K: C E=3: \frac{3}{2}: \frac{1}{2}: 1=6: 3: 1: 2$.

According to the Swallowtail Model, $S_{\triangle A D F}: S_{\triangle A E F}=D K: E K=3: 1$;

Noting that, $S_{\triangle A D F}=A F \times G I \div 2, \quad S_{\triangle A E F}=A F \times H J \div 2$;

Thus, $G I: H J=S_{\triangle A D F}: S_{\triangle A E F}=3: 1$;
That is, $G I \div H J=3$.