14. In an equilateral triangle $\mathrm{ABC}$, take any point $\mathbf{P}$ inside it, and draw perpendiculars $\mathrm{PD}, \mathrm{PE}$, and $\mathrm{PF}$ from this point to $\mathrm{BC}, \mathrm{CA}, \mathrm{AB}$ respectively, with $\mathrm{D}$, $\mathrm{E}$, and $\mathrm{F}$ being the feet of the perpendiculars. Calculate $\frac{P D+P E+P F}{B D+C E+A F}$.
14. Through point $\mathbf{P}$, draw three lines parallel to the three sides of a triangle (Figure 16). The three triangles formed (the shaded parts in the figure) are also equilateral triangles, and the perimeter of each of these triangles equals the side $A B=a$ of $\triangle A B C$. The sum of their heights equals the height of $\triangle \mathrm{ABC}$, thus
$$
\mathbf{P D}+\mathrm{PE}+\mathbf{P F}=\frac{\mathrm{a} \sqrt{3}}{2} .
$$

The sum of $B D+C E+A F$ equals the sides of the shaded triangles plus their halves, thus, $\mathrm{BD}+\mathbf{P E}+\mathrm{AF}=\frac{3}{2} \mathrm{a}$.

Therefore,
$$
\frac{P D+P E+P F}{B D+C E+A F}=\frac{1}{\sqrt{3}} \text {. }
$$