A Question related to triangle and centroid . The following is a geometry question I can't seem to get.
"Consider an acute angle △ABC. Points D, E, F are mid points of sides BC, CA and
AB respectively. G is the centroid of △ABC. Area of △AFG = 14, EC = 15/2. Perpendicular distance of F from BC = 6. Find
$BC^2 − AB^2$ "
Here is the figure that I have drawn -

From EC =15/2 I get AE=15/2 So AC =15 .
But what now? How can I utilize the fact that FK=6 and Area(AFG)=14 in finding $BC^2-AB^2$ ?
 A: Note : I got some amazing hints from Blue ( see comments on question ) and only because of them I could solve the problem . I'm answering my own question here but I'll be very grateful if someone else posted a simpler way to compute the answer .
Consider triangles AFG and BFG . They have equal bases ( AF=BF , since F is midpoint. ) They also have equal corresponding heights ( Since they have a common vertex at G ) . So their areas are equal. Similarly areas of all the sub-triangles ( BGD , CGD ...) are equal. Thus they all have area of 14 . So the area of whole triangle ABC = 6 *14 ( since there are 6 sub-triangles ) = 84. 
Now draw AL perpendicular to BC . Triangle BFK ~ ( similar ) Triangle BAL . But AB/BF=2 thus AL/FK=2 and AL=2*6=12 . Now Area of triangle ABC = 1/2 * base BC * height AL .
But we already know that area is 84 , thus 84=1/2*BC*12 and BC = 14 .
In triangle ALC , by Pythagoras theorem $CL^2=15^2 - 14^2$ ( Given EC=15/2 thus AC =15 ) .
Thus CL=9. Then BL=BC-CL = 14-9=5 . Now in triangle ABL , by Pythagoras theorem $AB^2 =AL^2+BL^2=12^2+  5^2 $ Thus $AB= 13 $.
Now $BC^2-AB^2=196-169=27 .$
