Let the length of edge of the regular hexagon be ' $x$ ' each.

EP and FQ are perpendiculars drawn on AD , so that $$\mathrm{EF}=\mathrm{PQ}=\mathrm{x}$$

The measure of each angle of a regular hexagon is $$\frac{(6-2) \times 180}{6}=120^{\circ}$$ (The measure of each angle of any regular polygon having $$n$$ sides is $$\left.\frac{(n-2) \times 180}{n}\right)$$

So, the measure of angle DEP $$=$$ angle $$\mathrm{DEF}-$$ angle $$\mathrm{PEF}=120^{\circ}-90^{\circ}=30^{\circ}$$

So, the triangle EDP is a $$90-60-30$$ triangle in which the side opposite to 90 degree i.e. $$E D$$ is $$x$$, so the length of the side opposite to 30 degrees i.e. DP would be $$\frac{x}{2}$$ which is same as the length of QA (using symmetry)

As shown in the figure on the right in a triangle having angles 30,60
$$\& 90$$, the corresponding opposite sides are $$a, a \sqrt{3}$$\& 2 a$$ respectively.

Finally the length of $$\mathrm{DA}=\mathrm{DP}+\mathrm{PQ}+\mathrm{QA}=\frac{x}{2}+{x}+\frac{x}{2}=2 \mathrm{x}$$ which is same as $\$mathrm{AF}+\mathrm{FE}=\mathrm{x}+\mathrm{x}=$$ $2 x$

Hence column A quantity is same as column $$B$$, so the answer is (C).