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).

Using the law of cosines,

  |EG|^2 = |EH|^2 + |HG|^2 - 2|EH||HG|cos(EHG)

         = |EH|^2 + |HG|^2 - 2|EH||HG|cos(t+15)

         = |EF|^2 + |FG|^2 - 2|EF||FG|cos(EFG)

         = |EF|^2 + |FG|^2 - 2|EF||FG|cos(180-[t-x])

  |FH|^2 = |FE|^2 + |EH|^2 - 2|FE||EH|cos(FEH)

         = |FE|^2 + |EH|^2 - 2|FE||EH|cos(t-15)

         = |HG|^2 + (sqrt(3)+1)^2*|EH|^2

           - 2(sqrt(3)+1)*|HG||EH|cos(t-15)

         = |FG|^2 + |GH|^2 - 2|FG||GH|cos(FGH)

         = |FG|^2 + |GH|^2 - 2|FG||GH|cos(180-[t+x])

  ,where 2x = Angle EFG - Angle FGH.

Combining the above two equations we get

  cos(t-x) - cos(t+x) = 

     (sqrt(3)+1)[cos(t-15) - cos(t+15)]

          or

  sin(x) = (sqrt(3)+1)sin(15)

          or

  sin(x)^2 = (sqrt(3)+1)^2*sin(15)^2

           = 2(sqrt(3)+2)*[1-cos(30)]/2

           = 1/2

Therefore, Angle EFG - Angle FGH = 90 degrees.

Posted by Bractals on 2007-11-08 14:20:06