I'm trying to prove that any regular and convex 2n-polygon can be divided into rhombuses.

I proceed by induction.

BASE CASE: For n=2 this is obviously true because a square is a rhombus.

INDUCTIVE HYPOTHESIS: suppose the statement is true for a polygon with 2n sides.

Now to make a polygon with 2(n+1) sides, we take the 2n-polygon which has already been divided into rhombuses. We take any n sides and replace them with n+2 sides, thus 'elongating' the 2n-polygon (so to speak) in such a way that we draw n lines parallel to those n sides, at some distance. And then join them with the 2n-polygon with a line at each end. Hence n more rhombuses are added to the 2n-polygon and this new shape is a 2n+2-polygon.

This completes my proof.

I have a feeling there's some loophole in this argument :s Can someone kindly help?