 | William Mitchell Gillespie - 1855 - 436 σελίδες
...proposition of Geometry, that in any figure bounded by straight lines, the sum of all the interior angles is equal to twice as many right angles, as the figure has sides less two ; since the figure can be divided into that number of triangles. Hence this common rule. "... | |
 | Euclides - 1856 - 168 σελίδες
...straight lines from a point F within the figure to each of the angles. And by the last proposition all the angles of these triangles are equal to twice as many right angles as there are triangles or sides to the figure. And the same angles are equal to the internal angles of... | |
 | Cambridge univ, exam. papers - 1856 - 200 σελίδες
...superposition. 3. Prove that all the internal angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides; and that all the external angles are together equal to four right angles. In what sense are these propositions... | |
 | Henry James Castle - 1856 - 220 σελίδες
...angles are the exterior angles of an irregular polygon ; and as the sum of all the interior angles are equal to twice as many right angles, as the figure has sides, wanting four ; and as the sum of all the exterior, together with all the interior angles, are equal... | |
 | William Mitchell Gillespie - 1856 - 478 σελίδες
...proposition of Geometry, that in any figure bounded by straight lines, the sum of all the interior angles is equal to twice as many right angles, as the figure has sides less two ; since the figure can be divided into that number of triangles. Hence this common rule. "... | |
 | William Mitchell Gillespie - 1857 - 538 σελίδες
...proposition of Geometry, that in any figure bounded by straight lines, the sum of all the interior angles is equal to twice as many right angles, as the figure has sides less two ; since the figure can be divided into that number of triangles. Hence this common rule. "... | |
 | Elias Loomis - 1858 - 256 σελίδες
...F, that is, together with four right angles (Prop. V., Cor. 2). Therefore the angles of the polygon are equal to twice as many right angles as the figure has sides, wanting four right angles. Cor. 1. The sum of the angles of a quadrilateral is four right angles ;... | |
 | W. Davis Haskoll - 1858 - 422 σελίδες
...and in an irregular polygon they may be all unequal. The interior angles of a polygon are together equal to twice as many right angles as the figure has sides, less four. On this is based the theory of the traverse, of which further explanation will be given... | |
 | Thomas Hunter - 1878 - 142 σελίδες
...other, the remaining angles must be equal. Cor. 2. The sum of all the interior angles of a polygon is equal to twice as many right angles as the figure has sides, minus four right angles. In the case of the triangle, this corollary has just been demonstrated; for,... | |
 | Āryabhaṭa - 1878 - 100 σελίδες
...been proved by the foregoing corollary, that all the interior angles together with four right angles are equal to twice as many right angles as the figure has sidesTherefore all the interior angles together with all the exterior angles are equal (Ax. 1) to all... | |
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... figure, together with four right angles, are equal to twice as many right angles as the figure has be divided into as many triangles as the figure has sides, by drawing straight lines from a point F within the figure to each of its angles. 
