EFGH is a convex quadrilateral having diagonals EG and FH. It is known that EF = GH, while EH/FG = √3 + 1 and Angle GHE - Angle HEF = 30^{o}.

Determine the value of (Angle EFG - Angle FGH).

Let the measure of angle HEF be denoted as e, EFG as f, FGH as g, and GHE as h.

Let x be the length of equal sides EF and GH.

Then using the Law of Cosines, four equations can be made:

(FH)^2 = (FG)^2 + x^2 - 2x*(FG)*cos(g)

(FH)^2 = (HE)^2 + x^2 - 2x*(HE)*cos(e)

(EG)^2 = (FG)^2 + x^2 - 2x*(FG)*cos(f)

(EG)^2 = (HE)^2 + x^2 - 2x*(HE)*cos(h)

The two equations for (FH)^2 can be combined to make:

[(HE)^2 - (FG)^2]/(2x) = (HE)*cos(e) - (FG)*cos(g)

Similarily for (EG)^2:

[(HE)^2 - (FG)^2]/(2x) = (HE)*cos(h) - (FG)*cos(f)

The last two equations can be combined to form:

(HE)*(cos(e) - cos(h)) = (FG)*(cos(g) - cos(f))

This equation can be simplified further by using the identity:

cos(a) - cos(b) = 2 * sin((b+a)/2) * sin((b-a)/2)

2*(HE) * sin((h+e)/2) * sin((h-e)/2) = 2*(FG) * sin((f+g)/2) * sin((f-g)/2)

Using the fact that e+f+g+h = 360 deg:

sin((f+g)/2) = sin(180 - (h+e)/2) = sin((h+e)/2)

Combining the last two equations and simplifying yields:

(HE) * sin((h-e)/2) = (FG) * sin((f-g)/2)

Now, solving the problem is a matter of substitution:

h-e = 30 deg, HE = 1+sqrt(3), FG = 1

sin(15) = (sqrt(6) - sqrt(2))/4

(1+sqrt(3)) * ((sqrt(6) - sqrt(2))/4) = (1) * sin((f-g)/2)

1/sqrt(2) = sin((f-g)/2)

f-g = 90 deg