Coordinate Geometry
Let the center of the hexagon = (0,0)
Let ED = 2
D= (1, sqrt 3)
F = (-2,0)
A= (-1 , -sqrt 3)
N = (-3/2 , -sqrt (3) / 2 )
M = (0, sqrt (3) )
B = (1 , -sqrt (3))
Slope of line containing segment (DN) = [ sqrt (3) + sqrt (3) / 2 ] / [5/2] = 3sqrt (3) / 5
Equation of this line
y = [ (3/5)sqrt 3 ] ( x - 1) + sqrt (3) (1)
Slope of line containing segment MB = [ 2sqrt (3) ] / -1 = -2sqrt (3)
Equation of this line
y = [ -2sqrt (3) ] (x ) + sqrt (3) (2)
Set (1) = (2) to find the x coordinate of P
[ (3/5) sqrt (3) ] (x -1) + sqrt (3) = [-2sqrt (3)]x + sqrt (3)
(3/5)sqrt (3) x - (3/5)sqrt (3) = -2sqrt (3)x
(13/5) sqrt (3) x = (3/5)sqrt (3)
(13/5)x = (3/5)
x = (3/5)(5/13) = 15/65 = 3/13
y = -2sqrt (3) (3/13) + sqrt (3)
y = (7/13)sqrt (3)
DP^2 = ( 1-3/13)^2 + ( sqrt 3 - (7/13)sqrt 3)^2
DP^2 = (10/13)^2 + [(6/13) sqrt 3]^2
DP^2 = 16/13
PN^2 = ( 3/13 + 3/2)^2 + [ (7/13)sqrt 3 + sqrt (3) / 2)^2
PN^2 = (45/26)^2 + (27sqrt (3) / 26)^2
PN^2 = 81/13
DP^2 / PN^2 = 16 / 81
DP / PN = 4 / 9