Each exterior angle of a regular hexagon is 60° (360° / 6), so when sides HB and IC are extended to meet at A , an equilateral triangle, ABC is created. Let the sides of this triangle be of length x. As BC, DE and FG are all parallel, triangles ABC, ADE and AFG are all equilateral. So, $<\mathrm{span}\mathrm{style}="\mathrm{font}-\mathrm{style}:\mathrm{italic};»\mathrm{DE}</\mathrm{span}>=<\mathrm{span}\mathrm{style}="\mathrm{font}-\mathrm{style}:\mathrm{italic};»\mathrm{DA}</\mathrm{span}>=p+x$; and $<\mathrm{span}\mathrm{style}="\mathrm{font}-\mathrm{style}:\mathrm{italic};»\mathrm{FG}</\mathrm{span}>=<\mathrm{span}\mathrm{style}="\mathrm{font}-\mathrm{style}:\mathrm{italic};»\mathrm{FA}</\mathrm{span}>=q+p+x$.