It is well known that, given two simple n-sided polygons, it may not be possible to triangulate the two polygons in a compatible fashion, if one's choice of triangulation vertices is restricted to polygon corners. Is it always possible to produce compatible triangulations if additional vertices inside the polygon are allowed? We give a positive answer and construct a pair of such triangu-lations with O(n 2) new triangulation vertices. Moreover, we show that there exists a \universal" way of triangulating an n-sided polygon with O(n 2) extra triangulation vertices. Finally, we also show that creating compatible triangulations requires a quadratic number of extra vertices in the worst case. 1 The Problem Given two simple polygons P 1 and P 2 , each with n vertices, it is not always possible to nd compatible triangulations of the two polygons. In other words, there may not exist a circular labeling of the corners of each polygon by consecutive numbers 1 through n and a set of n ? 3 non-crossing interior chords in P 1 , so that the open chords joining corresponding vertices in P 2 are disjoint and lie completely inside P 2. Consider gure 1|here each hexagon admits a unique triangulation, but the two triangulations are not compatible in the above sense, since the three interior chords \fan out" from a common vertex in P 1 and form a triangle in P 2. The following question was asked by Goodman and Pollack G]: Is it possible to triangulate any two polygons with the same number of vertices compatibly if additional vertices (Steiner points) are allowed in the interior? If yes, how many such points are required?