Question 1183719
Consider hexagon ABCDEF with the vertices labeled clockwise.  From the hint, line GH would be parallel to both lines CD and AF.  

The distance between lines CD and AF is {{{sqrt(3)*a}}}.  (Why is this so?)

Incidentally, since line GH is three-fourth of this distance away from line AF, the distance between lines GH and AF is {{{(3/4)*sqrt(3)*a}}}.

Let P be the intersection of lines FG and AH.
Let Q be the midpoint of line AF, and let R be the midpoint of line GH.
Then the points P, Q, R, and the center O are all collinear.


===> Triangle GRP is similar to triangle FQP.  
We know that {{{abs(GR) = (3a)/4}}} and {{{abs(FQ) = a/2}}}.  Now let {{{abs(QP) = alpha}}}, so that {{{abs(PR) = (3/4)*sqrt(3)*a - alpha}}}.

By similarity of triangles, 
{{{abs(GR)/abs(PR) = abs(FQ)/abs(QP)}}}, i.e.,

{{{((3a)/4)/((3/4)*sqrt(3)a - alpha) = (a/2)/alpha}}}

===> {{{alpha = abs(QP) = (3sqrt(3)*a)/10}}} after solving for {{{alpha}}}.

This implies that segment PO has length {{{(sqrt(3)/2)*a - (3sqrt(3)*a)/10 = (sqrt(3)/5)*a}}}.

Also, {{{abs(PR) = abs(QR) - abs(QP) = (3/4)*(sqrt(3)a) - (3sqrt(3)*a)/10 = ((9*sqrt(3))/20)*a}}}.


If we let K be the intersection of lines BE and FG, the triangle KOP will be similar to triangle GRP.  Then 

{{{abs(OK)/( (sqrt(3)/5)*a) = ((3a)/4)/(((9*sqrt(3))/20)*a))}}}


===> {{{abs(OK) = a/3}}}.


But {{{abs(OK)}}} is half of {{{abs(JK)}}}.  Therefore,

{{{highlight(abs(JK)) = 2*abs(OK) = highlight((2/3)*a)}}}.