Calculating Angles at Vertices 
This is quite a tricky question and I can't answer it.

Ok, I finally solved it and can prove the sum is 180 degrees. 
Let's mark the vertices clockwise, from A to G. I will refer to A, B .. G as angles of the 'star'.
We form and take as example AED triangle; we know the sum of all angles is 180. Thus A + ADE + DEA = A + E + BED + EDG + D = 180  ( 1 ). 
Let be M the point of intersection between BE and GD. In EMD triangle, we have angle EMD = 180 - BED - EDG ( 2 ).
From ( 1 ) and ( 2 ) => A + E + D = EMD ( 3 ). (similar procedure here for others)
Let be N the point of intersection between AD and FC. Let be P the point of intersection between AE and FC. In ANP triangle we have sum of all angles: A + APN + ANP = 180 ( 4 ).
But actually, ANP = DNC. DNC = G + D + C as shown above in similar procedure as ( 3 ). The same: APN = FPE = F + B + E.
If we put all things together, formula ( 4 ) becomes A + B + C + D + E + F = 180 (Q.E.D)
