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Then to prove (c), note that the radius joining the origin to a vertex of the polygon ''P''<sub>''n'' − 1</sub> makes an angle of less than 2{{pi}}/3 with each of the edges of the polygon at that vertex if exactly two triangles of ''P''<sub>''n'' − 1</sub> meet at the vertex, since each has an angle less than {{pi}}/3 at that vertex. To check this is true when three triangles of ''P''<sub>''n'' − 1</sub> meet at the vertex, ''C'' say, suppose that the middle triangle has its base on a side ''AB'' of ''P''<sub>''n'' − 2</sub>. By induction the radii ''OA'' and ''OB'' makes angles of less than 2{{pi}}/3 with the edge ''AB''. In this case the region in the sector between the radii ''OA'' and ''OB'' outside the edge ''AB'' is convex as the intersection of three convex regions. By induction the angles at ''A'' and ''B'' are greater than {{pi}}/3. Thus the geodesics to ''C'' from ''A'' and ''B'' start off in the region; using the Klein model, the triangle ''ABC'' lies wholly inside the region. The quadrilateral ''OACB'' has all its angles less than {{pi}} (since ''OAB'' is a geodesic triangle), so is convex. Hence the radius ''OC'' lies inside the angle of the triangle ''ABC'' near ''C''. Thus the angles between ''OC'' and the two edges of ''P''<sub>''n'' – 1</sub> meeting at ''C'' are less than {{pi}}/3 + {{pi}}/3 = 2{{pi}}/3, as claimed.
To prove (b), it must be checked how new triangles in ''P''<sub>''n''</sub> intersect. First consider the tiles added to the edges of ''P''<sub>''n'' – 1</sub>.
By convexity the new triangle added on an edge of ''P''<sub>''n'' − 1</sub> lies inside a sector emanating from the origin and going through the endpoints of the edge. Since these open sectors are disjoint, new triangles of this type only intersect as claimed.
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