25. The diagram shows two equilateral triangles. The distance from each point of the smaller triangle to the nearest point of the larger triangle is $\sqrt{3}$, as shown.
What is the difference between the lengths of the edges of the two triangles?
A $2 \sqrt{3}$
B $4 \frac{1}{2}$
C $3 \sqrt{3}$
D 6
E $4 \sqrt{3}$
25. D In the diagram shown, $F$ is the foot of the perpendicular from $D$ to $A B, G$ is the foot of the perpendicular from $D$ to $A C$ and $H$ is the foot of the perpendicular from $E$ to $A B$. So $D F=D G=E H=\sqrt{3}$. Consider triangles $A F D$ and $A G D$. Edge $A D$ is common to both triangles, $D F=D G$ and angles $A F D$ and $A G D$ are both right angles. So the triangles are congruent (RHS).
Therefore $\angle D A F=\angle D A G=60^{\circ} \div 2=30^{\circ}$.
So in triangle $A F D, \tan 30^{\circ}=\frac{D F}{A F}$.

Therefore $A F=\sqrt{3} \div(1 / \sqrt{3})=\sqrt{3} \times \sqrt{3}=3$. The same argument shows that $B H=3$. It is clear that $F H=D E$, so $A B-D E=A F+B F=3+3=6$.
Therefore the difference between the lengths of the edges of the two triangles equals 6 .