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Indeed as in the first step of the process there are two types of tile in building ''P''<sub>''n''</sub> from ''P''<sub>''n'' – 1</sub>, those attached to an edge of ''P''<sub>''n'' – 1</sub> and those attached to a single vertex. Similarly there are two types of vertex, one in which two new tiles meet and those in which three tiles meet. So provided that no tiles overlap the previous argument shows that angles at vertices do not exceed {{pi}} and implies that ''P''<sub>''n''</sub> is a convex polygon.
The equality above for ''a'', ''b'' and ''c''
The equality above for ''a'', ''b'' and ''c'' implies that if one of the angles is a right angle, say ''a'' = 2, then both ''b'' and ''c'' are greater than 2 and one of them, ''b'' say, must be greater than 3. In this case, reflecting the triangle across the side AB gives an isosceles hyperbolic triangle with angles {{pi}}/''c'', {{pi}}/''c'' and 2{{pi}}/''b''. The construction of the tessellation above through increasing convex polygons adapts word for word to this case except that around the vertex with angle 2{{pi}}/''b'', only ''b''—and not 2''b''—copies of the triangle are required to tile a neighborhood of the vertex. This is possible because the doubled triangle is isosceles.<ref>{{harvnb|Caratheodory|1954|pages=181–182}}</ref>▼
:<math>{1\over a} + {1\over b} + {1\over c} < 1</math>
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===Triangles with one or two cusps===
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