|AB| = |BC|, |CD| = |DE|,

   /ABC = 150°, /CDE = 30°,

   and |BD| = 2.

the pentagon can be broken down into 3 triangles, namely ABC CDE and ACE.  let x=|AB|=|BC|, y=|CD|=|DE|, and t=angle ACE.

now from law of cosines we have

|AC|^2=2*x^2*(1-Cos(150))=2*x^2*(1+Cos(30))    (1)

|CE|^2=2*y^2*(1-Cos(30))                       (2)

|BD|^2=4=x^2+y^2-2*x*y*Cos(t+90)     (3)

now the areas of the trinangles can be found using the area formula S1*S2*Sin(t) where s1,s2 are sides and t is the angle between them.

ACE=x^2*Sin(150)=x^2*Sin(30)

CDE=y^2*Sin(30)

ACE=AC*CE*Sin(t)

now AC*CE can be simplified using (2) and (3) as such

AC*CE=x*y*Sqrt(4*(1-cos(30))(1+cos(30)))

AC*CE=2xySqrt(1-cos^2(30))

AC*CE=2xySin(30)

so ACE=2xySin(30)Sin(t)

now sin(t)=-cos(t+90) thus

ACE=-2xySin(30)Cos(t+90)

thus when we add up the 3 areas we can factor out Sin(30)/2 and get

AREA=(Sin(30)/2)*(x^2+y^2-2xyCos(t+90))

this is where we can eliminate x,y,t completely using (1) and thus prove that the area is constant regardless of x,y,t.

AREA=(Sin(30)/2)*4=2Sin(30)=2*(1/2)=1